# How do you use the chain rule to differentiate f(x)=sec^2(3x^6-6x+7)tan^2(16x^-2+61cos(x^2))?

Jun 2, 2017

sec^2(3x^6-6x+7)xx2tan(16x^(-2)+61cos(x^2))sec^2(16x^(-2)+61cos(x^2)))xx(-32x^(-3)-122xsin(x^2))+2tan(3x^6-6x+7)sec^2(3x^6-6x+7)xx(18x^5-6)xxtan^2(x)(16x^(-2)+61cos(x^2))

#### Explanation:

The easiest way to break down this complicated derivative is to use the chain rule and the product rule. However, before we do that, let's write down the list of different functions and their respective derivatives.

{:(color(red)(f_1(x)=sec^2(x)),color(blue)(f'_1(x)=2tan(x)sec^2(x))),(color(green)(f_2(x)=tan^2(x)),color(orange)(f'_2(x)=2tan(x)sec^2(x))),(color(brown)(f_3(x)=3x^6-6x+7),color(black)(f'_3(x)=18x^5-6)),(color(gray)(f_4(x)=16x^(-2)+61cos(x^2)), color(magenta)(f'_4(x)=-32x^(-3)-122xsin(x^2))):}

Using this method of labeling the different functions in the given question, we can write the original problem as:

$f \left(x\right) = \textcolor{red}{{f}_{1}} \left[\textcolor{b r o w n}{{f}_{3} \left(x\right)}\right] \times \textcolor{g r e e n}{{f}_{2}} \left[\textcolor{g r a y}{{f}_{4} \left(x\right)}\right]$

The Product Rule allows us to say that $\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right]$ is

$\textcolor{red}{{f}_{1}} \left(\textcolor{b r o w n}{{f}_{3} \left(x\right)}\right) \times \frac{d}{\mathrm{dx}} \left(\textcolor{g r e e n}{{f}_{2}} \left(\textcolor{g r a y}{{f}_{4} \left(x\right)}\right)\right) + \frac{d}{\mathrm{dx}} \left(\textcolor{red}{{f}_{1}} \left(\textcolor{b r o w n}{{f}_{3} \left(x\right)}\right)\right) \times \textcolor{g r e e n}{{f}_{2}} \left(\textcolor{g r a y}{{f}_{4} \left(x\right)}\right)$

The Chain Rule allows us to further break down the derivatives in the above product rule step.

$\frac{d}{\mathrm{dx}} \left(\textcolor{g r e e n}{{f}_{2}} \left(\textcolor{g r a y}{{f}_{4} \left(x\right)}\right)\right) = \textcolor{\mathmr{and} a n \ge}{f {'}_{2}} \left(\textcolor{g r a y}{{f}_{4} \left(x\right)}\right) \times \textcolor{m a \ge n t a}{f {'}_{4} \left(x\right)}$

and

$\frac{d}{\mathrm{dx}} \left(\textcolor{red}{{f}_{1}} \left(\textcolor{b r o w n}{{f}_{3} \left(x\right)}\right)\right) = \textcolor{b l u e}{f {'}_{1}} \left(\textcolor{b r o w n}{{f}_{3} \left(x\right)}\right) \times \textcolor{b l a c k}{f {'}_{3} \left(x\right)}$

Therefore the entire derivative of $\frac{d}{\mathrm{dx}} \left[f \left(x\right)\right]$ is

$\textcolor{red}{{f}_{1}} \left(\textcolor{b r o w n}{{f}_{3} \left(x\right)}\right) \times \left[\textcolor{\mathmr{and} a n \ge}{f {'}_{2}} \left(\textcolor{g r a y}{{f}_{4} \left(x\right)}\right) \times \textcolor{m a \ge n t a}{f {'}_{4} \left(x\right)}\right] + \left[\textcolor{b l u e}{f {'}_{1}} \left(\textcolor{b r o w n}{{f}_{3} \left(x\right)}\right) \times \textcolor{b l a c k}{f {'}_{3} \left(x\right)}\right] \times \textcolor{g r e e n}{{f}_{2}} \left(\textcolor{g r a y}{{f}_{4} \left(x\right)}\right)$

The last step is plugging the functions and derivatives from the colorful table above:

color(red)(sec^2)(color(brown)(3x^6-6x+7))xxcolor(orange)(2tan(color(gray)(16x^(-2)+61cos(x^2)))sec^2(color(gray)(16x^(-2)+61cos(x^2))))xx(color(magenta)(-32x^(-3)-122xsin(x^2)))+color(blue)(2tan(color(brown)(3x^6-6x+7))sec^2(color(brown)(3x^6-6x+7))xx(color(black)(18x^5-6))xxcolor(green)(tan^2(x)(color(gray)(16x^(-2)+61cos(x^2))))