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

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Question:

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

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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)

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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.}

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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}

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©Ankitraj707



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