Question

given y = f(u) and u = g(x), find $\frac{dy}{dx}=f(g(x))g(x)$. y = 3u^8, u = 7x - 2 $\frac{dy}{dx}=square$

Explanation:

Step1: Find $f'(u)$

Differentiate $y = f(u)=3u^{8}$ with respect to $u$. Using the power - rule $\frac{d}{du}(au^{n})=nau^{n - 1}$, we have $f'(u)=3\times8u^{7}=24u^{7}$.

Step2: Find $g'(x)$

Differentiate $u = g(x)=7x - 2$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax + b)=a$, we get $g'(x)=7$.

Step3: Apply the chain - rule

By the chain - rule $\frac{dy}{dx}=f'(g(x))g'(x)$. Substitute $u = g(x)=7x - 2$ into $f'(u)$ and multiply by $g'(x)$. So $f'(g(x))=24(7x - 2)^{7}$, and $\frac{dy}{dx}=24(7x - 2)^{7}\times7$.

Answer:

$168(7x - 2)^{7}$