Question

Need to find the derivative of sin(tan(5*t^2/(8*t-3*t^3))). Please include all steps please.

Answer 1

Here is the equation once again.

Answer 2

This pretty much breaks down to the chain rule and quotient rule:

Sin(tan(5t^2/(8t - 3t^3)))

First: What is the derivative of Sin(x)? That's right! It's Cos(x)! but we need to chain rule it by taking the derivative of the tan(x) inside

Cos(tan(5t^2/(8t - 3t^2))) d/dt (tan(5t^2/(8t - 3t^2))

What is the derivative if tan(x)? It's just Sec^2(x) right, so we can put that in for the tan(x), but remember we have to chain rule the (5t^2)/(8t - 3t^2) so:

Cos(tan(5t^2/(8t - 3t^2))) Sec^2(5t^2/(8t - 3t^2)) d/dt (5t^2)/(8t - 3t^2)

What's the derivative of (5t^2)/(8t - 3t^2)? well we have to quotient rule and get

d/dt (5t^2)/(8t - 3t^2) = (8t - 3t^2)(10t) - (5t^2)(8 - 6t) / (8t - 3t^2)^2

= 80t^2 - 30t^3 - 40t^2 + 30t^3 / (8t - 3t^2)^2

= 40t^2/(8t - 3t^2)^2

so lets plug that back in for d/dt (5t^2)/(8t - 3t^2)

and we get:

Cos(tan(5t^2)/(8t - 3t^2))) Sec^2(5t^2/(8t - 3t^2)) (40t^2/(8t - 3t^2)^2)

tada!

Answer 3

Hermeezey... U have used (5t^2)/(8t - 3t^2)... it is 3t^3 not 2. look at my equation in Comment 1

Answer 4

haha sorry dude i messed up. This is what happens when you try to rush through Calculus. But you get the gist of how to do it right? I mean at least the steps to get the answer?

Answer 5

Yeh I do. Thanx mate. Just wanna check if I get the answer of 0 or not

Answer 6

I guess Maple simplifies the answer to 0. Thanks alot Hermeezey

Answer 7

Yeaaaa I'm not sure how to turn it to 0. I dumped most of my Calc 1 stuff for Calc 2.

Answer 8

The steps are obviously correct but how do you get a zero, thats the question... What do you think I should do? Your steps are obviously correct.. Any way to check if the answer to the derivative is correct or not?

Answer 9

Well I would say just to Integrate it, but I'm not sure if I could even do that. There really isn't a way to check. At least not one that I'm aware of.