Question 10.35 • Calculus (4th Edition) • Textbookish

\begin{align*} \frac{dz}{dx}&=\frac{dz}{dy}\cdot\frac{dy}{dx} \\ &=\cos y\cdot(1+2x) \\ &=\cos(x+x^2)(1+2x). \end{align*}

ii)

\begin{align*} \frac{dz}{dx}&=\frac{dz}{dy}\cdot\frac{dy}{dx} \\ &=\cos y\cdot(-\sin x) \\ &=-\cos(\cos x)\sin x. \end{align*}

iii)

\frac{dz}{dx}=\frac{dz}{du}\cdot\frac{du}{dx}=\cos u\cdot\cos x=\cos(\sin x)\cos x.

iv)

\begin{align*} \frac{dz}{dx}&=\frac{dz}{dv}\cdot\frac{dv}{du}\cdot\frac{du}{dx} \\ &=\cos v\cdot(-\sin u)\cdot\cos x \\ &=\cos(\cos(\sin x))(-\sin(\sin x))\cos x. \end{align*}

Solutions