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

Jan 28, 2017

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

Explanation:

$f ' \left(x\right) = \frac{1}{2 \sqrt{{\left(2 x - 5\right)}^{5} / {\left({x}^{2} + 2\right)}^{2}}} \textcolor{red}{\cdot \frac{5 \cdot {\left(2 x - 5\right)}^{4} \cdot 2 \cdot {\left({x}^{2} + 2\right)}^{2} - {\left(2 x - 5\right)}^{5} \cdot 2 \cdot \left({x}^{2} + 2\right) \cdot 2 x}{{x}^{2} + 2} ^ 4}$

$= \frac{1}{2} \sqrt{\frac{{\left({x}^{2} + 2\right)}^{2}}{2 x - 5} ^ 5} \cdot \frac{10 \cdot {\left(2 x - 5\right)}^{4} \cdot {\left({x}^{2} + 2\right)}^{2} - 4 x \cdot {\left(2 x - 5\right)}^{5} \cdot \left({x}^{2} + 2\right)}{{x}^{2} + 2} ^ 4$

$= \frac{1}{2} \sqrt{\frac{{\left({x}^{2} + 2\right)}^{2}}{2 x - 5} ^ 5} \cdot \frac{2 {\left(2 x - 5\right)}^{4} \left({x}^{2} + 2\right) \left(5 \left({x}^{2} + 2\right) - 2 x \left(2 x - 5\right)\right)}{{x}^{2} + 2} ^ 4$

=1/cancel2sqrt(((x^2+2)^2)/(2x-5)^5)*(cancel2 (2x-5)^4 cancel((x^2+2))(5x^2+10-4x^2+10x))/(x^2+2)^(cancel4^3

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

=cancel((x^2+2))/(cancel((2x-5)^2)sqrt(2x-5))*((2x-5)^(cancel4^2)(x^2+10x+10))/(x^2+2)^(cancel3^2

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