Question

sin³x cos⁵x,Find the derivative of the given function defined on proper domains.

Answer 1

it is given that function, f(x) = sin³x.cos⁵x
we have to find derivative of the given function.

f(x) = sin³x.cos⁵x

differentiate with respect to x,

df(x)/dx = d[sin³x.cos⁵x]/dx

=sin³x.d[cos⁵x]/dx + cos⁵x.d[sin³x]/dx

= sin³x.5cos⁴x .d[cosx]/dx + cos⁵x.3sin²x.d[sinx]/dx

= 5sin³x.cos⁴x.(-sinx) +3cos⁵x.sin²x.(cosx)

= -5sin⁴x.cos⁴x +3cos^6x.sin²x

= cos⁴x.sin²x [ 3cos²x - 5sin²x]

hence, first derivative of the given function is cos⁴x.sin²x [ 3cos²x - 5sin²x]

Answer 2

Hello,

Solution:
In this function we must apply chain rule,as follows

$\frac{d(uv)}{dx} = v \frac{du}{dx} + u \frac{dv}{dx}$

let V = sin³x

and U = cos⁵x

$\frac{d( {sin}^{3}x \: {cos}^{5} x )}{dx} = {sin}^{3}x \frac{d({cos}^{5} x)}{dx} + {cos}^{5} x \frac{d( {sin}^{3}x)}{dx} \\ \\ = {sin}^{3} x \: (5 \: {cos}^{4} x \frac{d( \: cos \: x)}{dx} ) \\ \\ + {cos}^{5} x \: (3 \: {sin}^{2} x \frac{d(sin \: x)}{dx} ) \\ \\ = {sin}^{3} x \: (5 \: {cos}^{4} x ( - \sin(x) ) \\ \\ + {cos}^{5} x \: (3 \: {sin}^{2} x (cos \: x))\\ \\ = - 5 {sin}^{4} x \: {cos}^{4} x + 3 {sin}^{2} x \: {cos}^{5} x$
Is the derivative of given function.

You can simplify this answer further as required.

Hope it helps you.