Q. 40 - Calculus

Question

Find the derivatives of the functions in Exercises 21–46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating. 

$f\left(x\right)=\frac{\left(3x+1\right){\left(x^4-3\right)}^4}{{\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5}$

Step-by-Step Solution

Answer

The required answer is $\frac{\left(3{\left(x^4-3\right)}^4+16x^3\left(3x+1\right){\left(x^4-3\right)}^3\right){\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5-(3x+1){\left(x^4-3\right)}^4\left(-2{\left(x+5\right)}^{-3}{\left(1+x^2\right)}^5+10x{\left(x+5\right)}^{-2}{\left(1+x^2\right)}^4\right)}{{\left(x+5\right)}^{-1}{\left(1+x^2\right)}^{{\displaystyle \frac{5}{2}}}}$

Step 1. Given Information

The given function is $f\left(x\right)=\frac{\left(3x+1\right){\left(x^4-3\right)}^4}{{\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5}$

Step 2. Calculation

Differentiate both the sides with respect to x, we get,      

$$\begin{gathered} f\text{'}\left(x\right)=\frac{\left({\displaystyle \frac{d}{dx}}\left(3x+1\right){\left(x^4-3\right)}^4\right){\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5-\left(3x+1\right){\left(x^4-3\right)}^4\left({\displaystyle \frac{d}{dx}}{\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5\right)}{{\left({\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5\right)}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{\left({\displaystyle 3{\left(x^4-3\right)}^4+\left(3x+1\right)}\right)\left(4{\left(x^4-3\right)}^3{\displaystyle \frac{d}{dx}}\left(x^4-3\right)\right){\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5-\left(3x+1\right){\left(x^4-3\right)}^4\left({\displaystyle -2{(x+5)}^{-2-1}\frac{d}{dx}\left(x+5\right)}\right){\left(1+x^2\right)}^5+{(x+5)}^{-2}\left(5{\left(1+x^2\right)}^4{\displaystyle \frac{d}{dx}}(1+x^2)\right)}{{\left({\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5\right)}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{\left({\displaystyle 3{\left(x^4-3\right)}^4+\left(3x+1\right)}\right)\left(4{\left(x^4-3\right)}^3\left(4x^3\right)\right){\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5-\left(3x+1\right){\left(x^4-3\right)}^4\left({\displaystyle -2{(x+5)}^{-3}}\right){\left(1+x^2\right)}^5+{(x+5)}^{-2}\left(5{\left(1+x^2\right)}^4\left(2x\right)\right)}{{\left({\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5\right)}^{{\displaystyle \frac{1}{2}}}} \\ =\frac{\left(3{\left(x^4-3\right)}^4+16x^3\left(3x+1\right){\left(x^4-3\right)}^3\right){\left(x+5\right)}^{-2}{\left(1+x^2\right)}^5-(3x+1){\left(x^4-3\right)}^4\left(-2{\left(x+5\right)}^{-3}{\left(1+x^2\right)}^5+10x{\left(x+5\right)}^{-2}{\left(1+x^2\right)}^4\right)}{{\left(x+5\right)}^{-1}{\left(1+x^2\right)}^{{\displaystyle \frac{5}{2}}}} \end{gathered}$$