Question

find second-degree derivative function y=(ax+b)m

Answer 1

HERE IS YOUR ANSWER

Given that :
$y = {(ax + b)}^{m}$

Now, differentiating with respect to x, we get :

$\frac{dy}{dx} = \frac{d}{dx} {(ax + b)}^{m} \\ \\ = m. {(ax + b)}^{m - 1} . \frac{d}{dx} (ax + b) \\ \\ = am {(ax + b)}^{m - 1}$

Again, differentiating with respect to x, we get :

$\frac{ {d}^{2} y}{d {x}^{2} } = \frac{d}{dx} am( {ax + b})^{m - 1} \\ \\ = am(m - 1) {(ax + b)}^{m - 2} \frac{d}{dx} (ax + b) \\ \\ = {a}^{2} m(m - 1) {(ax + b)}^{m - 2}$
which is the required 2nd degree derivative function.

RULE :

$\frac{d}{dx} ( {x}^{n} ) \\ \\ = n. {x}^{n - 1}$

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