Find $y'$ if $$y=\left( \left( 3x^2+2\right)^4+1\right)^5$$
The chain rule for differentiation is really quite simple. You just differentiate the 'outside function' first. Then multiply by the derivative of the 'inside function'. Easy peasy :-)
$\displaystyle y'=5\left( \left( 3x^2+2\right)^4+1\right)^4\times 4\left( 3x^2+2\right)^3\times 6x$
$\displaystyle y'=120x\left( 3x^2+2\right)^3\left( \left( 3x^2+2\right)^4+1\right)^4$
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