Question

Find the derrivative of y= cosx^3.sin^2(x^5) ​

Answer 1

Question:

Find the derivative of $y=cosx^3.sin^2(x^5)$

$\huge \underline{\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \:}$

We will use product rule.

Let $cosx^3$ be u and $sin^2(x^5)$ be v.

Now, the question is:

Differentiate y=uv.

y'= uv' + vu' (product rule)

$\huge \underline{\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

Let's differentiate u and v separately,we will later put the values.

$\sf u'= (cosx^3)' \\\\\sf \implies (x^3)' (-sinx^3) \\\\\sf \implies \color{blue}- sinx^3. 3x^2 \\\\\sf \boxed{Done}$

$\huge \underline{\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

Now, come to v'

$\sf v'= (sin^2(x^5))' can~ be ~written ~as ((sinx^5)^2)' \\\\\sf \implies 2sinx^5. (sinx^5)' \\\\\sf \implies 2sinx^5.cosx^5.(x^5)' \\\\\sf \implies 2sinx^5.cosx^5. 5x^4\\\\\sf \implies2.5x^4.sinx^5.cosx^5. \\\\\sf \implies \color{red} 10x^4.sinx^5.cosx^5 \\\\\sf \boxed{And ~it's~ done.}$

$\huge \underline{\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

Now, just put the values in y'= uv' + vu'

$\sf y'= cosx^3( \color{red}{10x^4.sinx^5.cosx^5} ) \color{black} +sin^2(x^5)( \color{blue}{-sinx^3. 3x^2} ) \\\\\sf \implies \boxed{ y'=10x^4.cosx^3.cosx^5.sinx^5- sin^2x^5.sinx^3.3x^2}$

$\huge \underline{\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:}$

©Ankitraj707