Just wanted to check if I am differentiating these functions correctly.

If a function can be simplified even further, can you please show me.

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- June 5th 2009, 12:31 AMdwatDifferentiation.
Just wanted to check if I am differentiating these functions correctly.

If a function can be simplified even further, can you please show me. - June 5th 2009, 12:44 AMmatheagle
- June 5th 2009, 12:45 AMRandom Variable
1)

2) correct

3) correct

4) - June 5th 2009, 01:05 AMdwat
- June 5th 2009, 01:08 AMmatheagle
you're having trouble with the chain rule

as I wrote above you CANNOT change the inner function g(x)

In the outer function is ln and the inner is cosine

The derivative of the log is 1 over the ENTIRE inner function giving us

Next multiply that by the derivative of the inner function

the derivative of the cosine x is .

Next multiply these two. - June 5th 2009, 01:11 AMdwat
- June 5th 2009, 01:15 AMmatheagle
It's f'(g(x))=1/cos x times g'(x) which is -sin x.

Better now? - June 5th 2009, 01:26 AMdwat
I didn't change the inner function, what I did was I only differentiated which is where I got then left the inner function as and then differentiate the inner function to . Hence where I came up with

But I see how you did it. 1 over the inner function left untouch. I'm struggling to understand why ends up within the differentiation of unless you replace the in the function with the inner function.

So basically, any would produce a solution with where the inner function will always end up at the bottom replacing e.g.

is that correct?

BTW, thank you for your help - June 5th 2009, 07:04 AMmatheagle
you did change the inner function, which is cosine x not just x.

The chain rule is just the product of two functions f' of g(x) which is one over g(x)=cos x here, then times the derivative of cos x.

Try setting f(x)=ln x and g(x) =cos x. Obtain the derivatives and plug into f'(g(x)) and g'(x).

and yes, the derivative of ln sin x is cot x.

More importantly you now know the integral of tan x and cot x.