Question

find the derivative of sin⁵(ax-c)​

Answer 1

We have to find out the derivative of sin⁵(ax-c).

We will differentiate by using Chain rule.

$\rm \longrightarrow \: \dfrac{d \bigg({sin}^{5} (ax - c) \bigg)}{dx} \\ \\ \\ \rm \longrightarrow 5 \: {sin}^{4} (ax - c)\: \times \dfrac{d \bigg({sin} (ax - c) \bigg)}{dx} \\ \\ \\ \rm \longrightarrow 5 \: {sin}^{4} (ax - c)\: \times \: cos (ax - c) \times \dfrac{d (ax - c) }{dx} \\ \\ \\ \rm \longrightarrow 5 \: {sin}^{4} (ax - c)\: \times \: cos (ax - c) \times \bigg(\dfrac{d (ax ) }{dx} - \dfrac{d (c ) }{dx}\bigg)\\ \\ \\ \rm \longrightarrow 5 \: {sin}^{4} (ax - c)\: \times \: cos (ax - c) \times \bigg(\dfrac{a \: d (x ) }{dx} - 0\bigg)\\ \\ \\ \rm \longrightarrow 5 \: {sin}^{4} (ax - c)\: \times \: cos (ax - c) \times a \\ \\ \\ \rm \longrightarrow 5 a\: \: {sin}^{4} (ax - c)\: \: cos (ax - c)$

Additional Information :

d(e^x)/dx = e^x

d(x^n)/dx = n x^(n-1)

d(ln x)/dx = 1/x

d(sin x)/dx = cos x

d(cos x)/dx = - sin x

d(tan x)/dx = sec² x

d(sec x)/dx = tan x * sec x

d(cot x)/dx = - cosec²x

d(cosec x)/dx = - cosec x * cot x