Q.If $y = x$ and $z = \dfrac{1}{x}$, then $\dfrac{dy}{dz} =$

For $y=e^{2x}$, the second derivative at $x=0$ is $4$.

Differentiate using the chain rule (the derivative of $e^{u}$ is $e^{u}\cdot u'$):

$\frac{dy}{dx}=e^{2x}\cdot 2 = 2e^{2x}.$

Differentiate again: …

For $y=4ae^{4x}$, $\dfrac{dy}{dx}=16ae^{4x}$.

Here $4a$ is a constant multiplier and $e^{4x}$ is differentiated by the chain rule (derivative of $e^{u}$ is $e^{u}\cdot u'$, with $u=4x$, $u'=4$): …

Differentiate both with respect to $x$ and use $\tfrac{dy}{dz}=\tfrac{dy/dx}{dz/dx}=\tfrac{1}{-1/x^2}=-x^2$.

Both $y$ and $z$ are given as functions of $x$, so use the chain-rule ratio $\dfrac{dy}{dz}=\dfrac{dy/dx}{dz/dx}$.

$y = x \;\Rightarrow\; \frac{dy}{dx} = 1$ …

Differentiating $y=e^{2x}$ twice gives $y_1=2e^{2x}$ and then $y_2=4e^{2x}$.

Given $y = e^{2x}$. Here $y_2$ means the second derivative $\dfrac{d^2y}{dx^2}$.

First derivative (chain rule, derivative of the exponent $2x$ is $2$):

$y_1 = \frac{dy}{dx} = e^{2x}\cdot 2 = 2e^{2x}.$

…

Differentiating $y=x$ and $z=\tfrac{1}{x}$ with respect to $x$ and dividing gives $\dfrac{dy}{dz}=\dfrac{1}{-1/x^2}=-x^2$.

Given $y = x$ and $z = \dfrac{1}{x} = x^{-1}$.

Differentiate each with respect to $x$:

$\frac{dy}{dx} = 1, \qquad \frac{dz}{dx} = -x^{-2} = -\frac{1}{x^2}.$

…