Hi folks,

Given $y = e^x \tan x$ we have to show that $\dfrac{d^2 y}{dx^2} = e^x (2 \tan^3 x + 2 \tan^2 x + 3 \tan x + 2)$

now I get $\dfrac{d^2 y}{dx^2} = e^x (2 \sec^2 x \tan x + 2 \sec^2 x + \tan x)$

so somehow $2 \sec^2 x \tan x + 2 \sec^2 x + \tan x = 2 \tan^3 x + 2 \tan^2 x + 3 \tan x + 2$

I have been horsing around for ages not getting there. A hint would be nice, lest I die of old age.

Thanks

Originally Posted by s_ingram
Hi folks,

Given $y = e^x \tan x$ we have to show that $\dfrac{d^2 y}{dx^2} = e^x (2 \tan^3 x + 2 \tan^2 x + 3 \tan x + 2)$

now I get $\dfrac{d^2 y}{dx^2} = e^x (2 \sec^2 x \tan x + 2 \sec^2 x + \tan x)$

so somehow $2 \sec^2 x \tan x + 2 \sec^2 x + \tan x = 2 \tan^3 x + 2 \tan^2 x + 3 \tan x + 2$

I have been horsing around for ages not getting there. A hint would be nice, lest I die of old age.

Thanks
Hint: $\displaystyle tan^2 (x) + 1 = sec^2(x)$

-Dan

Have you thought about the alternative to "dying of old age"?

"Lest I die of old age ..... without getting an answer!" was what I meant, but you are right. I didn't say what I meant.
With Dan's hint it came out in a few seconds.
The only alternative I can think of to dying of old age, is to die young!