Question

given y = f(u) and u = g(x) below, find $\frac{dy}{dx}=f(g(x))g(x)$. y = - sin u, u = 4x + 7 cos x $\frac{dy}{dx}=square$

Explanation:

Step1: Find $f'(u)$

Differentiate $y =-\sin u$ with respect to $u$. Using the derivative formula $\frac{d}{du}(\sin u)=\cos u$, we get $f'(u)=-\cos u$.

Step2: Find $g'(x)$

Differentiate $u = 4x + 7\cos x$ with respect to $x$. Using the sum - rule of differentiation and the derivative formulas $\frac{d}{dx}(x)=1$ and $\frac{d}{dx}(\cos x)=-\sin x$, we have $g'(x)=\frac{d}{dx}(4x)+\frac{d}{dx}(7\cos x)=4-7\sin x$.

Step3: Apply the chain - rule

Substitute $u = g(x)=4x + 7\cos x$ into $f'(u)$ and then multiply by $g'(x)$. So $\frac{dy}{dx}=f'(g(x))g'(x)=-\cos(4x + 7\cos x)(4 - 7\sin x)$.

Answer:

$-(4 - 7\sin x)\cos(4x + 7\cos x)$