Trigometric function of a function question.

• Jun 12th 2012, 09:35 AM
imonlyhuman
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. :)
• Jun 12th 2012, 09:43 AM
Reckoner
Re: Trigometric function of a function question.
Quote:

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,

$\frac{dy}{dx} = 42u^5\cos x = 42\sin^5x\cos x$
• Jun 12th 2012, 09:53 AM
imonlyhuman
Re: Trigometric function of a function question.

Is this the last step?

Apologies, trigonometric substitution confuses me!
• Jun 12th 2012, 10:03 AM
Reckoner
Re: Trigometric function of a function question.
Quote:

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 $x$, 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,

$y = 7\sin^6x$

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

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