Question

Derivatives of sinusoidal functions help?

y=secx/ cos^2x

I tried it but my answers is different than the back? Help

Answer 1

\[y=\frac{\sec x}{\cos^2x}=\sec x\cdot\sec^2x=\sec^3x\]
Try power/chain rule on that. Also, what are you getting? It could be that you're off by some rewriting using an identity or two.

Answer 2

I got secxtanx/ -2sinxcosx

Answer 3

And what's the answer in the back of the book?

Answer 4

3tanx(cosx)^-3

Answer 5

Okay, I'm guessing you tried using the quotient rule, correct? However, you didn't use it correctly.
If \(y=\dfrac{\sec x}{\cos^2x}\), then
\[y'=\frac{(\sec x)'\cdot\cos^2x - \sec x\cdot(\cos^2x)'}{\left(\cos^2x\right)^2}\]

Answer 6

I think I know what to do now. I use the quotient rule, then simplify, then bring the denominator to the top. Sorry, its just really confusing doing derivatives with trig. I could recognize it that well. I just learned it a day ago.

Answer 7

one more question. I did the quotient rule, but Now i have secxtanxcosx+ 2sinxsecx on the top and cos^3x on the bottom, how do i get it to match the book?

Answer 8

no need for quotient rule. use the form like @sithsandgiggles used
\[
y=\sec^3x\implies y'=3\sec^2x{d\over dx}(\sec x)\\
y'=3\sec^2x\sec x\tan x\implies y'=3\tan x(\sec x)^3\\
\boxed{y'=3\tan x(\cos x)^{-3}}
\]

Answer 9

i don't get how that form works though

Answer 10

\[\cos x={1\over\sec x}\]