Question

find the differentiation​

Answer 1

Answer:

the answer is

$- \frac{a}{2 \sqrt[3]{ax + b} }$

Step-by-step explanation:

As we know that

$\frac{d}{dx} \frac{u}{v} = \frac{v \frac{d}{dx }u - u \frac{d}{dx} v }{ {v}^{2} }$

Also ,

$\frac{d}{dx} { \sqrt{ax + b} } = \frac{d}{dx} {(ax + b)}^{ \frac{ 1 }{2 } } \\ = \frac{1}{2} \frac{d}{dx} {(ax + b)}^{ \frac{ 1 }{2 } - 1 } \\ = \frac{1}{2} {(ax + b)}^{ \frac{ - 1 }{2 } } \frac{d}{dx}(ax + b) \\ = \frac{1}{2 \sqrt{ax + b} } \times( a + 0) \\ = \frac{a}{2 \sqrt{ax + b} }$

$\frac{1}{ \sqrt{ax + b}} = \frac{\sqrt{ax + b} \: \frac{d}{dx} 1 - \: 1\frac{d}{dx} \: \sqrt{ax + b}}{ (\sqrt{ax + b})^{2} } \\ = \frac{\sqrt{ax + b} \times 0 \: - 1 \times \frac{1}{2 \sqrt{ax + b} } \: \frac{d}{dx} (ax + b)}{ax + b} \\ = \frac{0 - \frac{1}{2\sqrt{ax + b}} \: (a + 0) }{ax + b} \\ = \frac{ - \frac{ 1}{2\sqrt{ax + b}}a }{ax + b} \\ = - \frac{a}{2 \sqrt{ax + b}(ax + b) } \\ = - \frac{a}{2 {(ax + b)}^{ \frac{1}{2} }(ax + b) } \\ = - \frac{a}{2 {(ax + b)}^{ \frac{3}{2} } } \\ = - \frac{a}{2 \sqrt[3]{ax + b} }$