Question

given y = f(u) and u = g(x), find $\frac{dy}{dx}=f(g(x))g(x)$. y = $sqrt3{u^{2}}$, u = cos x $\frac{dy}{dx}=square$

Explanation:

Step1: Rewrite y in power - form

$y = u^{\frac{2}{3}}$

Step2: Differentiate y with respect to u

Using the power - rule $\frac{d}{du}(u^n)=nu^{n - 1}$, we have $y^\prime=f^\prime(u)=\frac{2}{3}u^{-\frac{1}{3}}$

Step3: Differentiate u with respect to x

$u = \cos x$, so $u^\prime=g^\prime(x)=-\sin x$

Step4: Apply the chain - rule $\frac{dy}{dx}=f^\prime(g(x))g^\prime(x)$

Substitute $u = \cos x$ into $f^\prime(u)$ and multiply by $g^\prime(x)$. We get $\frac{dy}{dx}=\frac{2}{3}(\cos x)^{-\frac{1}{3}}\times(-\sin x)=-\frac{2\sin x}{3\sqrt[3]{\cos x}}$

Answer:

$-\frac{2\sin x}{3\sqrt[3]{\cos x}}$