More Differentiation

a) i) $sec(x)tan(x)$

ii) $-cosec(x)cot(x)$

iii) $-cosec^{2}(x)$

b) $y=sec(cot(x))$

This is a function of a function, so use chain rule.

$dfrac{dy}{dx}=dfrac{dy}{dt}dfrac{dt}{dx}$

Take $t=cot(x)$

Then $y=sec(t)$

$dfrac{dy}{dt}=sec(t)tan(t)$

$dfrac{dt}{dx}=-cosec^{2}(x)$

$dfrac{dy}{dx}=-sec(t)tan(t)cosec^{2}(x)$

Put $x$ back in

$dfrac{dy}{dx}=-sec(cot(x))tan(cot(x))cosec^{2}(x)$

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This is a product of two functions, so use product rule.

$dfrac{dy}{dx}=udfrac{dv}{dx}+vdfrac{du}{dx}$

$u=sin(3x^{2});;;v=ln(x)$

$dfrac{dv}{dx}=dfrac{1}{x}$

For $dfrac{du}{dx}$ we need the chain rule because we have a function within a function.

$dfrac{du}{dx}=dfrac{du}{dt}dfrac{dt}{dx}$

Take $t=3x^{2}$

Then $y=sin(t)$

$dfrac{du}{dt}=cos(t)$

$dfrac{dt}{dx}=6x$

$dfrac{du}{dx}=6xcos(t)$

Put $x$ back in

$dfrac{du}{dx}=6xcos(3x^{2})$

Put into product rule:

$dfrac{dy}{dx}=dfrac{sin(3x^{2})}{x}+6xln(x)cos(3x^{2})$

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$y=9e^{frac{1}{2}x+cos(x)}$

First, find the derivative using the chain rule.

$dfrac{dy}{dx}=9left(dfrac{1}{2}-sin(x)right)e^{frac{1}{2}x+cos(x)}$

Stationary points are where this is equal to $0$.

$9left(dfrac{1}{2}-sin(x)right)e^{frac{1}{2}x+cos(x)}=0$

Exponentials can never be equal to $0$, so:

$9left(dfrac{1}{2}-sin(x)right)=0$

$dfrac{1}{2}-sin(x)=0$

$sin(x)=dfrac{1}{2}$

$x=dfrac{pi}{6},dfrac{5pi}{6}$

Find $y$-coordinate of stationary points:

$y=9e^{frac{1}{2}timesfrac{pi}{6}+cos(frac{pi}{6})}$

$y=9e^{frac{pi}{12}+frac{sqrt{3}}{2}}$

$y=9e^{frac{1}{2}timesfrac{5pi}{6}+cos(frac{5pi}{6})}$

$y=9e^{frac{5pi}{12}-frac{sqrt{3}}{2}}$

Stationary points are:

$left(dfrac{pi}{6},9e^{frac{pi}{12}+frac{sqrt{3}}{2}}right),left(dfrac{5pi}{6},9e^{frac{5pi}{12}-frac{sqrt{3}}{2}}right)$

Next, differentiate again using the product rule.

$u=9left(dfrac{1}{2}-sin(x)right)$

$v=e^{frac{1}{2}x+cos(x)}$

$dfrac{du}{dx}=-9cos(x)$

$dfrac{dv}{dx}=left(dfrac{1}{2}-sin(x)right)e^{frac{1}2{x}+cos(x)}$

$dfrac{d^{2}y}{dx^{2}}=9left(left(dfrac{1}{2}-sin(x)right)^{2}-cos(x)right)e^{frac{1}{2}x-cos(x)}$

Before substituting in $x=dfrac{pi}{6}$ and $x=dfrac{5pi}{6}$, we should first notice that the exponential part of the expression is positive for any $x$, so we need not keep it when trying to determine the sign of $dfrac{d^{2}y}{dx^{2}}$.

Want to determine the sign of $9left(left(dfrac{1}{2}-sin(x)right)^{2}-cos(x)right)$

Secondly, we notice that the $x$ values came from $dfrac{1}{2}-sin(x)=0$, so this part will remain $0$ when substituting the $x$ values in. We can also get rid of the factor of $9$ without changing the sign.

This means that all we need to do is find the sign of $-cos(x)$.

$-cosleft(dfrac{pi}{6}right)=-dfrac{sqrt{3}}{2}<0$ so this is a maximum.

$-cosleft(dfrac{5pi}{6}right)=dfrac{sqrt{3}}{2}>0$ so this is a minimum.

Hence:

$left(dfrac{pi}{6},9e^{frac{pi}{12}+frac{sqrt{3}}{2}}right)$ is a maximum.

$left(dfrac{5pi}{6},9e^{frac{5pi}{12}-frac{sqrt{3}}{2}}right)$ is a maximum.

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