Question

question
for the following equation, evaluate $\frac{dy}{dx}$ when $x = 1$.
$y = 5x^{5}+2x^{3}-3x^{2}$

Explanation:

Step1: Differentiate term - by - term

Using the power rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have:
$\frac{dy}{dx}=\frac{d}{dx}(5x^{5})+\frac{d}{dx}(2x^{3})-\frac{d}{dx}(3x^{2})$
$=5\times5x^{4}+2\times3x^{2}-3\times2x$
$ = 25x^{4}+6x^{2}-6x$

Step2: Substitute $x = 1$

Substitute $x = 1$ into $\frac{dy}{dx}$:
$\frac{dy}{dx}\big|_{x = 1}=25\times1^{4}+6\times1^{2}-6\times1$
$=25 + 6-6$
$=25$

Answer:

$25$