# How do you use the chain rule to differentiate f(x)=sin(x^2)/(x^4-3x)^4?

Nov 5, 2017

$f ' \left(x\right) = \frac{\left(2 {x}^{5} - 6 {x}^{2}\right) \cos \left({x}^{2}\right) - \left(16 {x}^{2} + 12\right) \sin \left({x}^{2}\right)}{{x}^{4} - 3 x} ^ 5$

#### Explanation:

This is a rational expression that will require us to use the quotient rule to differentiate. First let's rewrite the function as a product since the product rule is easier to use.

$f \left(x\right) = \frac{\sin \left({x}^{2}\right)}{{x}^{4} - 3 x} ^ 4$ $= \sin \left({x}^{2}\right) {\left({x}^{4} - 3 x\right)}^{-} 4$

Recall that for a function of the form $f \left(x\right) = a \cdot b$

The product rule for differentiation is as follows:
$f ' \left(x\right) = a ' b + a b '$

Also notice that in our function, $a$ and $b$ are both compositions. Recall that the derivative of a composition requires us to use the chain rule.

For a function of the form $f \left(x\right) = f \left(g \left(x\right)\right)$, the derivative is $f ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$

So, applying the product rule and chain rule on our function, we have
$f ' \left(x\right) = \left[\cos \left({x}^{2}\right) \cdot 2 x\right]$$\left[{\left({x}^{4} - 3 x\right)}^{-} 4\right]$+[sin(x^2)][-4(x^4-3x)^-5 (4x^3-3)]

$f ' \left(x\right) = \frac{2 x \cos \left({x}^{2}\right)}{{x}^{4} - 3 x} ^ 4$- $\frac{4 \left(4 {x}^{2} - 3\right) \sin \left({x}^{2}\right)}{{x}^{4} - 3 x} ^ 5$

$f ' \left(x\right) = \frac{{x}^{4} - 3 x}{{x}^{4} - 3 x}$$\frac{2 x \cos \left({x}^{2}\right)}{{x}^{4} - 3 x} ^ 4$- $\frac{4 \left(4 {x}^{2} - 3\right) \sin \left({x}^{2}\right)}{{x}^{4} - 3 x} ^ 5$

$f ' \left(x\right) = \frac{\left({x}^{4} - 3 x\right) 2 x \cos \left({x}^{2}\right) - 4 \left(4 {x}^{2} - 3\right) \sin \left({x}^{2}\right)}{{x}^{4} - 3 x} ^ 5$

$f ' \left(x\right) = \frac{\left(2 {x}^{5} - 6 {x}^{2}\right) \cos \left({x}^{2}\right) - \left(16 {x}^{2} + 12\right) \sin \left({x}^{2}\right)}{{x}^{4} - 3 x} ^ 5$