Question

Differentiation help/ *question below* will give medal

Answer 1

Can I have some help on this, please?

Answer 2

\[y=\tan(e^x-1)\]

Answer 3

So here you need to know chain rule.

Answer 4

What is the derivative of tan(x) first?

Answer 5

The derivative of tan(x) is sec^2x

Answer 6

\[y'=\sec^2(e^x-1) \cdot [(e^x-1)]'\]

Answer 7

So now you just need to differentiate that last part in [ ]

Answer 8

Differentiating that gives:\[e ^{x}\]

Answer 9

\[y'=\sec^2(e^x-1) \cdot e^x \]
Now to try to get what you are suppose to show
it looks like we need to know a certain trig identity

Answer 10

recall \[y=\tan(e^x-1) \]
\[y^2=\tan^2(e^x-1)\]

Answer 11

Do you think you know what trig identity we need?

Answer 12

sec^2x?

Answer 13

\[\text{ recall the trig identity with name pythagorean identity } \\ 1+\tan^2(x)=\sec^2(x) \]

Answer 14

\[1+\tan^2(e^x-1)=\sec^2(e^x-1)\]

Answer 15

Hmm, yes. How do I proceed from there now?

Answer 16

well isn't tan^2(e^x-1) equal to y^2?

Answer 17

yes

Answer 18

do you still need help?

Answer 19

No, I got it from here now. Thank you :)