Question #910e9

1 Answer
Feb 23, 2018

8sec^2(x)tan^3(x)+8sec^4(x)tan(x)

Explanation:

Here, we'll have to use the Chain Rule, Product Rule, and rules for differentiating trigonometric functions. Let's factor out the constant so the differentiation process is much cleaner:

4d/dx(sec^2(x)tan^2(x))

Using the product rule, we get:

4(d/dx(sec^2(x))tan^2(x)+sec^2(x)d/dxtan^2(x))

d/dxsec^2(x)=2sec(x)*d/dxsec(x)=2sec(x)sec(x)tan(x)

d/dxtan^2(x)=2tan(x)d/dxtan(x)=2tan(x)sec^2(x)

Plugging in the above derivatives, our solution becomes:

4(2sec(x)sec(x)tan(x)tan^2(x)+sec^2(x)2tan(x)sec^2(x))

Simplifying, we get:

4(2sec^2(x)tan^3(x)+2sec^4(x)tan(x))

8sec^2(x)tan^3(x)+8sec^4(x)tan(x)