Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(2x^3)=6x^2\)

Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)

Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)

What is the gradient of the straight line y = 5x ?

\(4\pi \) is a constant

The derivative of a constant is zero

Use the power rule

\(\large \frac{4}{3}\pi\) is a constant

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{1}{x^2})=\large\frac{\mathrm{d}}{\mathrm{d}x}(x^{-2})=-2x^{-3}\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{2}{x^3})=\large\frac{\mathrm{d}}{\mathrm{d}x}(2x^{-3})=-6x^{-4}\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{3}{\sqrt{x}})=\frac{\mathrm{d}}{\mathrm{d}x}(3x^{-\frac{1}{2}})=-\frac{3}{2}x^{-\frac{3}{2}}=-\frac{3}{2\sqrt{x^3}}\)

Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(x^3+x^2)=3x^2+2x\)

Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(2x^4-3x^2)=8x^3-6x\)

Use the power rule \(\large\frac{\mathrm{d}}{\mathrm{d}x}(ax^n)=nax^{n-1}\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\sqrt{x}-\frac{1}{x})=\large\frac{\mathrm{d}}{\mathrm{d}x}(x^\frac{1}{2}-x^{-1})=\frac{1}{2}x^{-\frac{1}{2}}+x^{-2}\)

There is no need to use the product rule for differentiation. Expand the brackets first

\(\large y=x(x^2-2)=x^3-2x\)

\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=3x^2-2\)

There is no need to use the product rule for differentiation. Expand the brackets first

\(\large f(x)=(x-1)(x+2)=x^2+2x-x-2=x^2+x-2\)

\(\large f'(x)=2x+1\)

There is no need to use the quotient rule. Simple before differentiating

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\frac{x^3-1}{x})=\large\frac{\mathrm{d}}{\mathrm{d}x}(x^2- \frac{1}{x})=\large\frac{\mathrm{d}}{\mathrm{d}x}(x^2-x^{-1})\\ =2x+x^{-2}=2x+\frac{1}{x}\\ =\frac{2x^3+1}{x^2}\)

\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-6x\)

When \(\large x = \frac{1}{2}\), \(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-6\times \frac{1}{2}=-3\)

Expand the brackets before differentiating

\(\large f(x)=x^2(2x-1)=2x^3-x^2\)

\(\large f'(x)=6x^2-2x\)

\(\large f'(-1)=6(-1)^2-2(-1)=6+2=8\)

There is no need to use the quotient rule for differentiation. Simplify before differentiating

\(\large y=\frac{x-4}{x^2}=\frac{1}{x}-\frac{4}{x^2}=x^{-1}-4x^{-2}\)

\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-x^{-2}+8x^{-3}\)

\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{x^2}+\frac{8}{x^3}\\ \)

When x = 1 , \(\large\frac{\mathrm{d}y}{\mathrm{d}x}=-\frac{1}{1}+\frac{8}{1}=7\)

\(\large f(x)=2x^2-x^{\frac{1}{2}}\)

\(\large f'(x)=4x-\frac{1}{2}x^{-\frac{1}{2}}=4x-\frac{1}{2\sqrt{x}}\)

\(\large f'(0.25)=4(0.25)-\frac{1}{2\sqrt{0.25}}=1-\frac{1}{2\times 0.5}=1-1=0\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\sin x)=\cos x\\ \large\frac{\mathrm{d}}{\mathrm{d}x}(\cos x)=-\sin x\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(3\sin x)=3\cos x\\ \large\frac{\mathrm{d}}{\mathrm{d}x}(2\cos x)=-2\sin x\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(-5\cos x)=5\sin x\\ \large\frac{\mathrm{d}}{\mathrm{d}x}(-3\sin x)=-3\cos x\)

\(\large f'(x) = 2\cos x \\ \large f'(\frac{\pi}{3}) = 2\cos (\frac{\pi}{3}) =2\times\frac{1}{2}=1\)

\(\large f'(x) = -2\sin x \\ \large f'(\frac{\pi}{a}) = -2\sin (\frac{\pi}{a})=-\sqrt{2}\)

\(\large\sin (\frac{\pi}{a})=\frac{\sqrt{2}}{2}\)

\(\large\sin (\frac{\pi}{4})=\frac{\sqrt{2}}{2}\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(e^x)=e^x\)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln x)=\frac{1}{x}\)

Using the laws of logarithms, we know that

\(\large\ln 2x=\ln 2+\ln x \)

\(\large\frac{\mathrm{d}}{\mathrm{d}x}(\ln 2+\ln x)=0+\frac{1}{x}\)

\(\large y=\frac{2e^x+1}{3}=\frac{2}{3}e^x+\frac{1}{3}\)

\(\large\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{3}e^x\)

when x = 0 , \(\large\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2}{3}e^0=\frac{2}{3} \times 1=\frac{2}{3}\)

\(\large f'(x) = 3e^x-2 \cos x\)

\(\large f'(0) = 3e^0-2 \cos 0=3\times 1-2\times 1=1\)

We can use the chain rule

\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)

= 8u³

=8(2x - 5)³

You can use the chain rule, but you should try to remember that \(\frac{d}{dx}(e^{ax})=ae^{ax}\)

There are 2 ways of thinking about this type of derivative.

• You can use the chain rule \(f'(x)=\frac{3}{3x}=\frac{1}{x}\)

or

• You can use the properties of logarithms

f(x) = ln3x = ln3 + lnx

\(f'(x)=0+\frac{1}{x}\)

Notice that we can re-write the question to make it easier

\(y =\frac{1}{e^{x^{3}}}=e^{-x^{3}}\)

We can use the chain rule

\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)

= e^u(-3x²)

=\(-3x^{2}e^{-x^{3}}\)

It helps if you think of the question like this

\( {1\over 4x}= {1\over 4} x^{-1}\)

Some people get integration and differentiation mixed up for logarithm functions

Note that \(\int { \frac { 1 }{ 4x } dx= } \frac { 1 }{ 4 } \int { \frac { 1 }{ x } dx= } \frac { 1 }{ 4 } lnx \)

Use the chain for this, or use the quick rule

\(\frac { d }{ dx } [f(x)]^{ n }=nf'(x)[f(x)]^{ n-1 }\)

\(\frac { d }{ dx }(lnx)^{3}=3{1\over x}(lnx)^{2}\)

Use laws of logarithms to transform this question into something easier

\(lnx^{3}=3lnx\)

You should try to remember that \(\frac { d }{ dx } (sinax)=acosax\)

We can use the chain rule

\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)

= cosu(2x)

= \(2x(cosx^{2})\)

We can use the chain rule

\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)

= \(\frac{1}{2}u^{-\frac{1}{2}}cosx\)

= \(\frac{1}{2}(sinx)^{-\frac{1}{2}}cosx\)

=\(\frac{cosx}{2\sqrt{sinx}}\)

We can use the chain rule

\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)

= e^u(-1)

=\(-{ e }^{ \pi -x }\)

\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)

=-sinu (2)

=\(-2sin(2x-\frac { \pi }{ 4 } )\)

We can use the chain rule

\(\frac { dy }{ dx } =\frac { dy }{ du } \times \frac { du }{ dx } \)

=sec²u (-2)

=\(-2{ sec }^{ 2 }(\pi -2x)\)

We need to use the product rule

\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)

\( \frac{dy}{dx}=3x\times10(2x-1)^{4} \ +\ (2x-1)^{5}\times 3\)

We can factorise this result. There is a common factor of \(3(2x-1)^4\)

\( \frac{dy}{dx}=3(2x-1)^{4}[10x+(2x-1)]\)

\( \frac{dy}{dx}=3(2x-1)^{4}(12x-1)\)

We need to use the product rule

\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)

\( \frac{dy}{dx}=sinx\times(-2sin2x) \ +cosx\times cos2x\)

We need to use the product rule

\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)

\( \frac{dy}{dx}=x^2 \times{1 \over x} \ +lnx\times 2x\)

= x + 2xlnx

We need to use the product rule

\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)

\( \frac{dy}{dx}=2x \times\frac{-x}{\sqrt{1-x^2}} \ +\sqrt{1-x^2}\times 2\)

\(=\frac{-2x^2}{\sqrt{1-x^2}} \ +2\sqrt{1-x^2}\)

\(=\frac{-2x^2}{\sqrt{1-x^2}} \ + \ \frac{2(1-x^2)}{\sqrt{1-x^2}}\)

\(=\frac{2-4x^2}{\sqrt{1-x^2}} \)

We need to use the quotient rule

\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { sinx\cdot e^x -e^x\cdot cosx }{( sinx)^{ 2 }} \)

\(=\frac { e^x(sinx - cosx) }{( sinx)^{ 2 }} \)

We need to use the quotient rule

\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { cos2x\cdot 3e^{3x} -e^{3x}\cdot (-2sin2x) }{ (cos2x)^{ 2 } }\)

\(\frac { dy }{ dx } =\frac { 3e^{3x}cos2x +2e^{3x}sin2x }{ (cos2x)^{ 2 } }\)

We need to use the quotient rule

\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { (3x-2)\cdot (2x) -(x^2+1)\cdot 3}{ (3x-2)^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { 6x^2-4x -3x^2-3}{ (3x-2)^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { 3x^2-4x -3}{ (3x-2)^{ 2 } } \)

We need to use the Quotient Rule

\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { \sqrt{1-x^2}\cdot 2-2x\cdot \frac{-x}{\sqrt{1-x^2}} }{ 1-x^2 }\)

\(\frac { dy }{ dx } =\frac { \frac{2(1-x^2)}{\sqrt{1-x^2}}+ \frac{2x^2}{\sqrt{1-x^2}} }{ 1-x^2 }\)

\(\frac { dy }{ dx } =\frac { \frac{2-2x^2+2x^2}{\sqrt{1-x^2}}}{ 1-x^2 }\)

\(\frac { dy }{ dx } =\frac{2}{ (1-x^2)^{\frac{3}{2}} }\)

We need to use the product rule

\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)

\(\frac { dy }{ dx } =e^{ 2x }\frac { 1 }{ x } +lnx\cdot 2e^{ 2x }\)

\(\frac { dy }{ dx } =e^{ 2x }(\frac { 1 }{ x } +2lnx)\)

We need to use the product rule

\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)

\(\frac { dy }{ dx } =e^{ x }\frac { 2 }{ x } + 2e^{ x }lnx\)

\(\frac { dy }{ dx } =\frac { 2e^x }{ x } + 2e^{ x }lnx\)

We need to use the Quotient Rule

\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { lnx\cdot 2e^{ 2x }-e^{ 2x }\cdot \frac { 1 }{ x } }{ (lnx)^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { e^{ 2x }(2lnx -\frac { 1 }{ x }) }{ (lnx)^{ 2 } } \)

We need to use the Quotient Rule

\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { ln2x\cdot 3e^{ 3x }-e^{ 3x }\cdot \frac { 1 }{ x } }{ (ln2x)^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { e^{ 3x }(3ln2x-\frac { 1 }{ x } ) }{ (ln2x)^{ 2 } } \)

We need to use the Quotient Rule

\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { e^{ 2x }\cdot \frac { 1 }{ x } -2e^{ 2x }\cdot ln3x }{ ({ e }^{ 2x })^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { e^{ 2x }(\frac { 1 }{ x } -2ln3x) }{ ({ e }^{ 2x })^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { \frac { 1 }{ x } -2ln3x }{ { e }^{ 2x } } \)

\(\frac { dy }{ dx } =\frac { \frac { 1 }{ x } -\frac { 2xln3x }{ x } }{ { e }^{ 2x } } \)

\(\frac { dy }{ dx } =\frac { 1-2xln3x }{ { xe }^{ 2x } } \)

We need to use the Quotient Rule

\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { { tanx }\cdot e^{ x }-e^{ x }sec²x }{ (tanx)^{ 2 } } \)

\(\frac { dy }{ dx } =\frac { { e^xtanx }-e^{ x }sec²x }{ (tanx)^{ 2 } } \)

\(f(x)=2x^{0.5}\)

\(f'(x)=2(0.5)x^{-0.5}=x^{-0.5}\)

\(f''(x)=-0.5x^{-1.5}\)

We need to use the Quotient Rule

\(f(x)=\frac { g(x) }{ h(x) } \\ f'(x)=\frac { h(x)\cdot g'(x)-g(x)\cdot h'(x) }{ { \left[ h(x) \right] }^{ 2 } } \)

\(f'(x)=\frac { (x²+1)(3x²)-{ x }^{ 3 }(2x) }{ { (x²+1) }^{ 2 } } \)

\(f'(-1)=\frac { ((-1)²+1)(3(-1)²)-{ (-1) }^{ 3 }(2(-1)) }{ {((-1)²+1 ) }^{ 2 } } \)

\(f'(-1)=\frac { (2)(3)-{ (-1) }(-2) }{ { (2) }^{ 2 } } \)

\(f'(-1)=\frac { 4 }{ 4 } \)

The local maximum of a function occurs when the gradient = 0

We need to use the Quotient Rule to find the gradient function

\(f(x)=\frac { g(x) }{ h(x) } \\ f'(x)=\frac { h(x)\cdot g'(x)-g(x)\cdot h'(x) }{ { \left[ h(x) \right] }^{ 2 } } \)

\(f'(x) =\frac { { x }\cdot \frac { 1 }{ x } -lnx }{ x² } \)

\(f'(x)=\frac { 1-lnx }{ x² } \)

Solve \(f'(x) = 0\)

\(\frac { 1-lnx }{ x² } =0\)

1 - lnx = 0 , since

lnx = 1

x = e

This is a related rates of change question. We use the chain rule

\(\large \frac{dA}{dt}=\frac{dA}{dx} \cdot \frac{dx}{dt}\)

The area of the square is given by the formula

\(\large A=x^2\)

We differentiate with respect to x to find \(\large \frac{dA}{dx}\)

\(\large \frac{dA}{dx}=2x\)

We substitute this into the chain rule and rearrange

\(\large \frac{dA}{dt}=2x \cdot \frac{dx}{dt}\)

\(\large \frac{\frac{dA}{dt}}{2x}= \frac{dx}{dt}\)

This is a related rates of change question. We use the chain rule

\(\large \frac{dV}{dt}=\frac{dV}{dr} \cdot \frac{dr}{dt}\)

The volume of a sphere is given by

\(\large V= \frac{4}{3}\pi r^3 \)

We differentiate with respect to r to find \(\large \frac{dV}{dr}\)

\(\large \frac{dV}{dr}=4\pi r^2\)

We substitute this into the chain rule

\(\large \frac{dV}{dt}=4\pi r^2 \cdot \frac{dr}{dt}\)

This is a related rates of change question. We use the chain rule

\(\large \frac{dC}{dt}=\frac{dC}{dr} \cdot \frac{dr}{dt}\)

The circumference of a circle is given by the formula

\(\large C=2 \pi r\)

We differentiate with respect to r to find \(\large \frac{dC}{dr}\)

\(\large \frac{dC}{dr}=2\pi\)

We substitute this into the chain rule

\(\large \frac{dC}{dt}=2\pi \cdot \frac{3}{\pi} =6\)

This is a related rates of change question. We use the chain rule

\(\large \frac{dA}{dt}=\frac{dA}{dr} \cdot \frac{dr}{dt}\)

The area of a circle is given by the formula

\(\large A=\pi r^2\)

We differentiate with respect to r to find \(\large \frac{dA}{dr}\)

\(\large \frac{dA}{dr}=2\pi r\)

When r = 5cm, \(\large \frac{dA}{dr}=10\pi \)

We substitute this into the chain rule

\(\large \frac{dA}{dt}=10\pi \cdot \frac{1}{\pi} =10\)

This is a related rates of change question. We use the chain rule

\(\large \frac{dV}{dt}=\frac{dV}{dx} \cdot \frac{dx}{dt}\)

The volume of the cube is given by the formula

\(\large V=x^3\)

We differentiate with respect to x to find \(\large \frac{dV}{dx}\)

\(\large \frac{dV}{dx}=3x^2\)

When x = 10, \(\large \frac{dV}{dx}=300\)

We substitute this into the chain rule

\(\large30=300 \cdot \frac{dx}{dt}\)

\(\large \frac{dx}{dt}\)= 0.1

This is a related rates of change question. We use the chain rule

\(\large \frac{dV}{dt}=\frac{dV}{dr} \cdot \frac{dr}{dt}\)

The volume of a sphere is given by \(\large V= \frac{4}{3}\pi r^3 \)

We differentiate with respect to r to find \(\large \frac{dV}{dr}\)

\(\large \frac{dV}{dr}=4\pi r^2\)

when the radius, r = 4cm , \(\large \frac{dV}{dr}=4\pi \times 4^2=64 \pi\)

We substitute this into the chain rule

\(\large160\pi=64\pi\cdot \frac{dr}{dt}\)

\(\large \frac{dr}{dt}=\frac{160\pi}{64\pi}=2.5\)

A horizontal tangent has a gradient = 0

You need to find the value of x for which \(\large f'(x)=0\)

Use your graphical display calculator to sketch the graph of the tangent to the function f at x = 1

The equation of the tangent is found to be y = 2.39x + 0.307

Hence this intersects the y axis when x = 0

To 3 significant figures, this is y = 0.307

Use your graphical display calculator to sketch the graph of the normal to the function f at x = 0

Plot f and the equation of the normal on the same axes and find the point of intersection

Make sure that your calculator is in radians mode

Notice that the domain for \(\large f'\) exists for \(0\le x\le1\)

The points of inflexion for the graph of f exist where \(\large f''(x)=0\)

You can plot the graph of \(\large f'\) and find the x coordinate of the first stationary point for \(\large f'\), in this case, a local maximum

Use your calculator to find the root of the equation f(x) = 0

Ensure that you find the correct root

\(\large f(x)=\ln (x)-3x\ , \ x>0\)

\(\large f'(x)=\frac{1}{x}-3\)

\(\large f''(x)=-\frac{1}{x^2}\)

Solve \(\large \frac{1}{x}-3=-\frac{1}{x^2}\)

You can do this by plotting the graphs of \(\large y=\frac{1}{x}-3\) and \(\large y=-\frac{1}{x^2}\) and finding the x coordinate of the point of intersection

Note that the answer only needs to be given to 2 s.f.

Use your graphical display calculator to sketch the graph of the tangent to the function f at x = 1 and also for x = 2

Plot these two tangents and find the point of intersection

To find a point of inflexion, we need to solve \(\large f''(x)=0\)

This is also a stationary point of the function \(\large f'(x)\)

We can differentiate the function to find \(\large f'(x)=4x^3-2e^x\)

The curve of a function is concave up when \(\large f''(x)>0\)

This will occur between two points of inflexion

Differentiate \(\large f(x)=sin(e^x) \) using the Chain Rule

\(\large f'(x)=e^xcos(e^x) \)

We can find the points of inflexion for \(f(x)\) by finding the stationary points for \(f'(x)\)

You do not need to plot a graph of the curve and the tangents, but it may help you visualise the problem.

Plot the graph of the function and the tangent to the curve at the point x = 1

To get a true picture of the graph, you may want to plot the graph so that the axes are 1:1 (so that a right angle actually looks like a right angle).

You will notice that the gradient of this tangent = 2

You need to find the x coordinates of the graph for which the curve has a gradient = \(-\frac{1}{2}\)

Differentiate the function using the product rule

\(\large f'(x)=2 \ln x+2 \)

Solve \(\large 2 \ln x+2 =-\frac{1}{2}\)

You can use the equation solver or a graphical approach.

Here is what the two tangents look like

For a local maxima

• \(f'(x)=0\)
• \(f''(x)<0\)

At local maxima \(f'(x)\)= 0 and \(f''(x)<0\)

At local minima \(f'(x)\)= 0 and \(f''(x)>0\)

Note that, gradient at c is negative and gradient at e is positive. Therefore, \(f'(c)

The diagram below shows the graph of the function y = f(x) in black

The diagram below shows the graph of the function \(y = f''(x)\) in green

We are given the graph of the gradient function and asked to work out what the original function might look like

If there is a point of inflexion then \(f''(x)=0\)

If it is a non-stationary point then \(f'(x)\neq 0\)

The function must either be

• increasing from x=a to x=c, so \(f(a)<f(b)<f(c)\)<><>
• decreasing from x=a to x=c, so \(f(a)>f(b)>f(c)\)

The sign of the gradient at x=a and x=c must be the same (they must be either both positive, or both negative).

Differentiate \(y = x^3-2x^2\)

\(\frac{dy}{dx}=3x^2-4x\)

when x = 1, \(\frac{dy}{dx}=3(1)^2-4(1)=-1\)

y = sin3x,

\(\frac{dy}{dx}=3cos3x\)

We know that \(\frac{dy}{dx}=\frac{3}{2}\)

\(3cos3x=\frac{3}{2}\)

\(cos3x=\frac{1}{2}\)

Arccos(0.5)=\(\frac{\pi}{3}\)

3x = \(\frac{\pi}{3}\)

x = \(\frac{\pi}{9}\)

Differentiate \(y = \frac{e^x}{x^2}\)

We need to use the Quotient Rule

​\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)​

\(\frac{dy}{dx}=\frac { x^{ 2 }e^{ x }-e^{ x }\cdot 2x }{ (x^{ 2 })^{ 2 } } \)

when x = 1 , \(\frac{dy}{dx}=\frac { 1^{ 2 }e^{ 1 }-2e^{ 1 } }{ 1^{ 4 } } \)

\(\frac{dy}{dx}=\frac { e-2e }{ 1 } =-e\)

Gradient of tangent = -e

Gradient of normal = \(\frac{-1}{-e}=\frac{1}{e}\)

\(f(x)=e^{-x}\)

\(f'(x)=-e^{-x}\)

\(f'(x)= -\frac{1}{e^x}\)

\(f'(ln2)= -\frac{1}{e^{ln2}}=-\frac{1}{2}\)

To differentiate y = xtanx, we need to use the Product Rule

​\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)​

\(\frac{dy}{dx}=xsec²x+tanx\cdot1\)

When x = \(\pi\), \(\frac{dy}{dx}=\pi sec²(\pi)+tan(\pi)\)

\(\frac{dy}{dx}=\frac{\pi}{[ cosx(\pi)]^2}+tan(\pi)\)

\(\frac{dy}{dx}=\frac{\pi}{( -1)^2}+0=\pi\)

\(\frac{y-0}{x-\pi}=\pi\)

\(y=\pi(x-\pi)\)

\(y=\pi x-\pi^2\)

We need to use the product rule to differentiate \(y=x^3lnx\)

​\(y=uv\\ \frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}\)​

\(\frac{dy}{dx}=x^3\cdot \frac{1}{x}+lnx\cdot3x^2\)

\(\frac{dy}{dx}=x^2+3x^2lnx\)

\(\frac{dy}{dx}=x^2(1+lnx)\)

When x = 1, \(\frac{dy}{dx}=1(1+ln1)=1(1+0)=1\)

Gradient of tangent = 1

Gradient of normal = -1

Find the equation of the straight line with gradient = -1 which passes through the point (1,0)

\(\frac{y-0}{x-1}=-1\)

y = -x + 1

\(y=e^{sin2x}\)

\(\frac{dy}{dx}=2cos2x\cdot e^{sin2x}\)

When x = \(\frac{\pi}{2}\), \(\frac{dy}{dx}=2cos(\pi)\cdot e^{sin(\pi)}\)

\(\frac{dy}{dx}=2(-1))\cdot e^{0}=-2\)

Gradient of tangent = -2

Gradient of normal = \(\frac{-1}{-2}=\frac{1}{2}\)

Find the equation of the straight line with gradient = \(\frac{1}{2}\) which passes through the point \((\frac{\pi}{2},1)\)

\(\frac{y-1}{x-\frac{\pi}{2}}=\frac{1}{2}\)

\(2y-2=x-\frac{\pi}{2}\)

\(4y-4=2x-\pi\)

\(2x-4y+(4-\pi)=0\)

Find the gradient function using the Quotient Rule

​\(y=\frac { u }{ v } \\ \frac { dy }{ dx } =\frac { v\cdot \frac { du }{ dx } -u\cdot \frac { dv }{ dx } }{ v^{ 2 } } \)​

\(\frac{dy}{dx}=\frac { x\cdot\frac{1}{x}-lnx\cdot 1 }{ x^{ 2 } } \)

\(\frac{dy}{dx}=\frac { 1-lnx }{ x^{ 2 } } \)

Maximum value occurs where \(\frac{dy}{dx}=0 \)

Solve \(\frac { 1-lnx }{ x^{ 2 } } =0\)

Since \(x\neq 0\) , 1-lnx = 0

lnx = 1

x = e

Differentiate the function f(x) = tanx - 4x

\(f'(x)=sec^2x-4\)

Minimum value occurs when \(f'(x)=0\)

sec²x - 4 = 0

sec²x = 4

cos²x = \(\frac{1}{4}\)

cosx = \(\pm \frac { 1 }{ 2 } \)

\(Arccos(\frac{1}{2})=\frac{\pi}{3}\)

There is only one correct value in the interval \(0 <x<\frac { \pi }{ 2 } \)

Hence \(x=\frac{\pi}{3}\)

Use the Product Rule to find \(\frac{dy}{dx}\)

\(\frac{dy}{dx}=e^x+xe^x\)

Differentiate again to find \(\frac{d²y}{dx²}\)

\(\frac{d²y}{dx²}=e^x+e^x+xe^x\)

\(\frac{d²y}{dx²}=e^x(2+x)\)

Point of inflexion exists where \(\frac{d²y}{dx²}=0\)

\(e^x(2+x)=0\)

\(e^x=0\quad,\quad2+x=0\)

not possible , x = -2

The initial displacement is given by s(0)

s(0) = 3sin(0) - 4cos(0)

s(0) = -4(1) = -4

Velocity is given by the rate of change of displacement. We need to differentiate displacement:

s(t) = sin4t - cos3t

\(v(t)=s'(t) = 4cos4t + 3sin3t\)

Acceleration is the rate of change of velocity. We need to differentiate velocity

\(v(t) = 10e^{-0.5t}\)

\(a(t) = v'(t) = 10(-0.5)e^{-0.5t}\)

\(a(t) = -5e^{-0.5t}\)

Initial acceleration is a(0)

\(a(0) = -5e^{-0.5(0)}\)

\(a(0) = -5e^{0}=-5\)

Velocity is given by the rate of change of displacement. We need to differentiate displacement:

s(t) = 1 - cos2t

\(v(t)=s'(t) = 2sin2t\)

Maximum velocity occurs when \(v'(t)=0\)

\(v'(t) = 4cos2t\)

Solve 4cos2t = 0

cos2t = 0

Arccos(0) = \(\frac{\pi}{2}\)

\(2t=\frac{\pi}{2}\)

\(t=\frac{\pi}{4}\)

Acceleration is the rate of change of velocity.

Velocity is given by the rate of change of displacement.

We need to differentiate two times:

\(s(t) = e^{-2t}\)

\(v(t)=s'(t) = -2e^{-2t}\)

\(a(t)=v'(t)=s''(t) = 4e^{-2t}\)

The ball reaches its maximum height when velocity = 0

To find velocity we need to differentiate the equation for displacement (height):

\(s(t) = bt - 5t^2\)

\(v(t) = s'(t) = b- 10t\)

v(t) = 0

b - 10t = 0

b = 10t

Since t = 2.5,

b= 10x2.5 = 25

The object changes direction when s(t) is a maximum or minimum

\(s(t) = 2t^3-11t^2+16t-7\)

\(s'(t) = 6t^2-22t+16\)

Solve \(s'(t) = 0\)

\(6t^2-22t+16=0\\ 3t^2-11t+8=0\)

(3t - 8)(t - 1) = 0

\(t = \frac{8}{3}\quad,\quad t=1\)

Let's check that s(1) is a maximum or minimum

\(s'(t) = 6t^2-22t+16\)

\(s''(t) = 12t-22\)

\(s''(1) = 12-22<0\) Maximum

The object first changes direction when t = 1

The minimum velocity occurs when acceleration = 0

\(s(t) = sint+cost\)

\(v(t)=s'(t) = cost-sint\)

\(a(t)=v'(t)=s''(t) = -sint-cost\)

a(t) = 0

-sint - cost = 0

sint = -cost

\(\frac{sint}{cost}=-1\)

tant = -1

\(t = \frac{3\pi}{4}\)

Minimum velocity \(v( \frac{3\pi}{4}) = cos( \frac{3\pi}{4})-sin( \frac{3\pi}{4})\)

\(v( \frac{3\pi}{4}) = -\frac{1}{\sqrt{2}} -\frac{1}{\sqrt{2}}\)

\(=-\frac{2}{\sqrt{2}}\\=-\sqrt{2}\)