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Patience and careful substitution is all that is necessary to differentiate these types of functions.

Take the function: $y=4x^2(5x^3+2)^4$

First label the two separate parts which would be differentiated with the product rule, namely:

$u=4x^2\\;\\;\\;\\text{and}\\;\\;\\;v=(5x^3+2)^4$

We also know that, by the product rule:

$\\frac{dy}{dx}=u\\frac{dv}{dx}+v\\frac{du}{dx}$

Looking at the substitution for $u$, the derivate can be easily found with the power rule:

$\\frac{du}{dx}=(4\\times2)x^{2-1}=8x^1=8x$

So far, from the product rule, we have:

$\\frac{dy}{dx}=(4x^2)\\frac{dv}{dx}+(5x^3+2)^4(8x)$

Looking at the expression for $v=(5x^3+2)^4$ we can see that finding $\\frac{dv}{dx}$, however, requires the chain rule.

Our usual subsitution letter $u$ has already been assigned an expression in the product rule part. This time, we can use $t$.

By the chain rule:

$\\frac{dv}{dx}=\\frac{dv}{dt}\\times\\frac{dt}{dx}$

Let $t=5x^3+2$

This means that: $v=t^4\\;\\;\\;\\therefore\\;\\;\\;\\frac{dv}{dt}=4t^3$

Also, $t=5x^3+2\\;\\;\\;\\therefore\\;\\;\\;\\frac{dt}{dx}=15x^2$

It follows that: $\\frac{dv}{dx}=\\frac{dv}{dt}\\times\\frac{dt}{dx}=4t^3\\times15x^2=60x^2t^3$

Subsituting back the original expression for $t$:

$\\frac{dv}{dx}=60x^2(5x^3+2)^3$

We can now input our expression for $\\frac{dv}{dx}$ into the original product rule section.

$\\frac{dy}{dx}=(4x^2)\\frac{dv}{dx}+(5x^3+2)^4(8x)\\;\\;\\;\\therefore\\;\\;\\;\\frac{dy}{dx}=(4x^2)(60x^2(5x^3+2)^3)+(5x^3+2)^4(8x)$

This is simplified further:

$\\frac{dy}{dx}=240x^4(5x^3+2)^3+8x(5x^3+2)^4=8x(5x^3+2)^3\\left(30x^3+(5x^3+2)\\right)$

Finally:

$\\frac{dy}{dx}=8x(5x^3+2)^3(35x^3+2)$

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$\\simplify{{c[1]}x^3({c[2]}x+{c[3]})^{p[1]}}$

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$\\simplify{{c[4]}x^{p[2]}({c[5]}x^2+{c[6]})^{p[3]}}$

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Differentiate the following expressions with respect to $x$ using the both product rule and the chain rule.

Simplify your answers as much as possible.

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Using the chain rule within product rule problems.

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