Question

What is the derivative of f(x)=sec(x)(2x-tan(x))

Answer 1

You have 2 functions over there, so use the product rule.
e.g.
let \(f(x)=g(x)h(x)\)
then
\(\frac{df(x)}{dx}=g'(x)h(x)+g(x)h'(x)\)

Answer 2

Can you identify the 2 functions?

Answer 3

I don't think I can

Answer 4

Give it a try

Answer 5

Oh, would they be sec(x) and (2x-tan(x))

Answer 6

That's correct, now can you apply the product rule and tell me what you get

Answer 7

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

Answer 8

Yep :)

Answer 9

Sweet! But this is a multiple choice question and all the answers are over cos(x)^3. How would I change the answer to get it over cos(x)^3?

Answer 10

That's pretty easy, use the fact that sec(x)=1/cos(x) :D

Answer 11

\[\large -\frac{1}{\cos(x)} (-2+\frac{1}{\cos^2(x)}-2 x \frac{\sin(x)}{\cos(x)}(x)+\frac{\sin^2(x)}{\cos^2(x)})\]

Answer 12

hmmm... could you assist me with finding the final answer from that, please?

Answer 13

sure, the first step would be distributing -1/cos(x) to everything inside the parenthesis, careful with the negatives

Answer 14

\[\large \frac{2}{\cos(x)}-\frac{1}{\cos^3(x)}+\frac{2x\sin(x)}{\cos^2(x)}-\frac{\sin^2(x)}{\cos^3(x)}\]Then you add everything together
\[\large \frac{2\cos^2(x)}{\cos^3(x)}-\frac{1}{\cos^3(x)}+\frac{2x\sin(x)\cos(x)}{\cos^3(x)}-\frac{\sin^2(x)}{\cos^3(x)}\]\[\large\frac{2\cos^2(x)-1+2\sin(x)\cos(x)-\sin^2(x)}{\cos^3(x)}\]

Answer 15

Super!! Thank you so much! :)