Question

Find the derivative of the function f(x) = cos^2 (x) + tan^2 (x).

Answer 1

\[f \prime(x)=2\cos x(-\sin x)+...\]

Answer 2

um how did you get that?

Answer 3

ok. \(\large\color{black}{ \cos^2(x) }\) is same as, \(\large\color{black}{ (\cos x)^2 }\). Then he used the CHain rule.

Answer 4

\(\large\color{black}{ \frac{d}{dx}(\cos x)^2 =2(\cos x)^{2-1}\times(- \sin x) }\)

Answer 5

see?

Answer 6

so you just use the power rule and trig derivative? and the final answer would just be 2cosx x -sinx?

Answer 7

yes, you use the power rule for the cos(x), as if it was just x. THEN, you multiply times the derivative of the inner part. (The inner part is cos(x), and it's derivative is -sin(x). )

Answer 8

So, \(\large\color{black}{ \frac{d}{dx}(\cos x)^2 =2(\cos x)^{2-1}\times(- \sin x)=-2\cos x \sin x }\)

Answer 9

and you would do the same thing for tan?

Answer 10

yep you would apply this
\[\frac{d}{dx}(f(x))^{n}=n(f(x))^{n-1} \cdot f'(x) \]

Answer 11

then you would need the chain rule for \(\large\color{black}{ \tan^2(x) }\).

\(\large\color{black}{ \tan^2(x) =(\tan x)^2}\).

Answer 12

(yes, same thing for tan^2x)

Answer 13

ok so 2cosx(-sinx) x 2tanx(sec^2x)

Answer 14

you had a PLUS in between (inside the function), not a multipkication sign.

Answer 15

oh oops haha thank you again!

Answer 16

yes, so your answer (as you know, but just want to put it down), is

\(\large\color{black}{ f'(x)=-2\cos x \sin x+2\tan x \sec x }\)

Answer 17

you can re-write as,
\(\large\color{black}{ f'(x)=2\tan x \sec x-2\sin x \cos x }\)

Answer 18

What does your teacher require to do with the derivative, anything else you need? like re-write in cosines and sines?`

Answer 19

that is pretty much all she requires, thank you so much!

Answer 20

yw!