# How do you differentiate  f(x)= [(2x-5)^5]/[(x^2 +2)^2]  using the chain rule.?

##### 2 Answers
Jul 30, 2017

The derivative is $= \frac{2 {\left(2 x - 5\right)}^{4} \left({x}^{2} + 10 x + 10\right)}{{x}^{2} + 3} ^ 3$

#### Explanation:

This is the derivative of a quotient

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

Here,

$u \left(x\right) = {\left(2 x - 5\right)}^{5}$, $\implies$, $u ' \left(x\right) = 5 {\left(2 x - 5\right)}^{4} \cdot 2 = 10 {\left(2 x - 5\right)}^{4}$

$v \left(x\right) = {\left({x}^{2} + 2\right)}^{2}$, $\implies$, $v ' \left(x\right) = 2 \left({x}^{2} + 2\right) \cdot 2 x = 4 x \left({x}^{2} + 2\right)$

Therefore,

$f ' \left(x\right) = \frac{10 {\left(2 x - 5\right)}^{4} {\left({x}^{2} + 2\right)}^{2} - {\left(2 x - 5\right)}^{5} \left(4 x \left({x}^{2} + 2\right)\right)}{{x}^{2} + 2} ^ 4$

$= \frac{\left({x}^{2} + 2\right) {\left(2 x - 5\right)}^{4} \left(10 \left({x}^{2} + 2\right) - 4 x \left(2 x - 5\right)\right)}{{x}^{2} + 4} ^ 4$

$= \frac{{\left(2 x - 5\right)}^{4} \left(10 {x}^{2} + 20 - 8 {x}^{2} + 20 x\right)}{{x}^{2} + 2} ^ 3$

$= \frac{{\left(2 x - 5\right)}^{4} \left(2 {x}^{2} + 20 x + 20\right)}{{x}^{2} + 2} ^ 3$

$= \frac{2 {\left(2 x - 5\right)}^{4} \left({x}^{2} + 10 x + 10\right)}{{x}^{2} + 3} ^ 3$

Jul 30, 2017

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{2 {\left(2 x - 5\right)}^{4} \left({x}^{2} + 10 x + 10\right)}{{x}^{2} + 2} ^ 3$

#### Explanation:

First we need to use the quotient rule:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{\left({x}^{2} + 2\right)}^{2} \frac{d}{\mathrm{dx}} \left({\left(2 x - 5\right)}^{5}\right) - {\left(2 x - 5\right)}^{5} \frac{d}{\mathrm{dx}} \left({\left({x}^{2} + 2\right)}^{2}\right)}{{x}^{2} + 2} ^ 4$

Now using the chain rule:

$\frac{d}{\mathrm{dx}} \left({\left(2 x - 5\right)}^{5}\right) = 5 {\left(2 x - 5\right)}^{4} \frac{d}{\mathrm{dx}} \left(2 x - 5\right) = 10 {\left(2 x - 5\right)}^{4}$

d/dx ( (x^2+2)^2)) = 2(x^2+2)d/dx (x^2+2) = 4x(x^2+2)

so:

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{10 {\left({x}^{2} + 2\right)}^{2} {\left(2 x - 5\right)}^{4} - 4 x {\left(2 x - 5\right)}^{5} \left({x}^{2} + 2\right)}{{x}^{2} + 2} ^ 4$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{10 \left({x}^{2} + 2\right) {\left(2 x - 5\right)}^{4} - 4 x {\left(2 x - 5\right)}^{5}}{{x}^{2} + 2} ^ 3$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{\left(2 x - 5\right)}^{4} \left(10 \left({x}^{2} + 2\right) - 4 x \left(2 x - 5\right)\right)}{{x}^{2} + 2} ^ 3$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{\left(2 x - 5\right)}^{4} \left(10 {x}^{2} + 20 - 8 {x}^{2} + 20 x\right)}{{x}^{2} + 2} ^ 3$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{{\left(2 x - 5\right)}^{4} \left(2 {x}^{2} + 20 x + 20\right)}{{x}^{2} + 2} ^ 3$