This is a question about implicit differentiation.

You will need to find the y coordinate when x=-1 and then find \(\frac{dy}{dx} \) at this point.

To find \(\frac{dy}{dx} \) , differentiate \(x^2+2e^{(x+2y)}=3\) with respect to x.

From the Chain Rule, we know that \(\frac{d}{dx}(e^{f(x)})=f'(x)e^{f(x)}\)

Substitute x=-1 into equation

Find \(\frac{dy}{dx}\) by differentiating \( x^2+2e^{(x+2y)}=3\) with respect to x

From the Chain Rule, we know that \(\frac{d}{dx}(e^{f(x)})=f'(x)e^{f(x)}\)

\(2x+2(1+2\frac{dy}{dx})e^{(x+2y)}=0\)

Substitute x=-1, y=0.5

\(2(-1)+2(1+2\frac{dy}{dx})e^{(-1+2(0.5)}=0\\ -2+2(1+2\frac{dy}{dx})e^{(0)}=0\\ 2(1+2\frac{dy}{dx})=2\\ 1+2\frac{dy}{dx}=1\\ 2\frac{dy}{dx}=0\\ \frac{dy}{dx}=0 \)

This is a question about implicit differentiation.

Differentiate \(x\arcsin y=e^{2y}\) with respect to x and find the gradient function.

You might need to look up \(\frac{d}{dx}( arcsin x)\) in the information booklet.

You will need to use the Product Rule, since \(x\arcsin y\) is the product of two functions.

Careful rearrangement will be required!

Note that \(\frac{d}{dx}(\arcsin x)=\frac{1}{\sqrt{1-x^{2}}}\)

Therefore, \(\frac{d}{dx}(\arcsin y)=\frac{1}{\sqrt{1-y^{2}}}\frac{dy}{dx}\)

Differentiate \(x\arcsin y=e^{2y}\) with respect to x

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

Rearrange the equation to collect terms with \(\frac{dy}{dx}\) on one side

\(x\frac{1}{\sqrt{1-y^{2}}}\frac{dy}{dx}-2e^{2y}\frac{dy}{dx}=-\arcsin y\)

Multiply both sides by -1

\(2e^{2y}\frac{dy}{dx}-x\frac{1}{\sqrt{1-y^{2}}}\frac{dy}{dx}=\arcsin y\)

Factorise

\((2e^{2y}-\frac{x}{\sqrt{1-y^{2}}})\frac{dy}{dx}=\arcsin y\)

Divide both sides through by \(2e^{2y}-\frac{x}{\sqrt{1-y^{2}}}\)

\(\frac{dy}{dx}=\frac{\arcsin y}{2e^{2y}-\frac{x}{\sqrt{1-y^{2}}}}\)

Multiply numerator and denominator by \({\sqrt{1-y^{2}}}\)

\(\frac{dy}{dx}=\frac{\sqrt{1-y^{2}}\arcsin y}{2e^{2y}\sqrt{1-y^{2}}-x}\)

This is a question about implicit differentiation. If you need a reminder about stationary points visit stationary points page

a) Differentiate \(x^2y+2x=1\) with respect to x. The stationary point exists where \( \frac{dy}{dx}=0\). You can find an equation for y in terms of x by substituting this into your equation. Can you use this and the initial equation to find the coordinates?

b) Differentiate the equation for a second time. Remember that local minima occurs when \(\frac{d^2y}{dx^2}>0 \)

a)

Differentiate \(x^2y+2x=1\) with respect to x.

\(2xy + x^2\frac{dy}{dx}+2=0\)

Substitute \(\frac{dy}{dx}=0 \)

\(2xy + 0 + 2 = 0\)

\(xy = -1\)

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

Substitute \(y = \frac{-1}{x}\) into initial implicit equation.

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

Hence, a stationary point exists at (1, -1)

b)

\(2xy + x^2\frac{dy}{dx}+2=0\)

Differentiate \(2xy + x^2\frac{dy}{dx}+2=0\) with respect to x

\(2y + 2x\frac{dy}{dx}+ 2x\frac{dy}{dx}+ x^2\frac{d^2y}{dx^2}=0\)

Hence

\(2y + 4x\frac{dy}{dx}+ x^2\frac{d^2y}{dx^2}=0\)

Substitute x=1 , y= -1, \(\frac{dy}{dx}=0\)

\(-2 + 0 + \frac{d^2y}{dx^2}=0\\ \frac{d^2y}{dx^2}=2 \)

Since \(\frac{d^2y}{dx^2}>0\) , stationary point at (1, -1) is a local minima