Thread: Trigometric function of a function question.

Hi Guys,

I am having problems with the following question..

Evaluate dy(over)dx for y=7sin^6x

Here is where i am up to..

Let u = sinx > du(over)dx = cosx
y = 7u^6 > dy(over)du = 42u5
dy(over)dx = dy(over)du x du(over)dx = 42u^5x (cosx)

I now have to substitute U back?

Any help appreciated, thanks.

Originally Posted by imonlyhuman

Let u = sinx > du(over)dx = cosx
y = 7u^6 > dy(over)du = 42u5
dy(over)dx = dy(over)du x du(over)dx = 42u^5x (cosx)

I now have to substitute U back?

It's a little confusing when you use "x" as a multiplication sign and as a variable. But if I understand what you are saying, your work should be correct so far. Now you substitute,

$\displaystyle \frac{dy}{dx} = 42u^5\cos x = 42\sin^5x\cos x$

Thanks for your reply Reckoner,

Is this the last step?

Apologies, trigonometric substitution confuses me!

Originally Posted by imonlyhuman

Is this the last step?

Apologies, trigonometric substitution confuses me!

Yes, that is the last step. We want the derivative in terms of $\displaystyle x$, which we have.

As you get more comfortable using the Chain Rule, you may find it easier to differentiate functions of this sort without making a formal substitution. For example,

$\displaystyle y = 7\sin^6x$

$\displaystyle \frac{dy}{dx} = 42\sin^5x\cdot\frac d{dx}\left[\sin x\right]$

$\displaystyle =42\sin^5x\cos x$