Simplifying the Derivative of log(sqrt(1+log x)-sin x)

[courtrigrad]

Hello all:

I need help in finding the derivative of:

log( sqrt(1+ log x) - sin x )

I know that derivative of log x is 1/x.

1/ sqrt(1 + logx) - sin x ) = sqrt(1+log x) + sin x / ( 1 + log x - sin ^2 x)

Then I found derivative of inside expression and multiplied with the previous derivative. I get something almost the same as the answer, but I can't seem to simplify it.

The answer is:

(1- 2x)* sqrt(1 + log x cos x) / 2x * sqrt(1 + log x)* sqrt(1 + log x - sin x)

Any help is greatly appreciated

Thanks!

 

[mprm86]

Using the chain rule, if you make u = (sqrt(1+logx) - sinx), then the derivative will be
1/u * du/dx, where du/dx is:
1/2sqrt(1+logx) * 1/x - cosx

I hope this will help. When you are derivating sqrt(1+logx), you have to use again the cahin rule, so it will be sqrt(v), and the derivative will be:
1/2sqrt(v) * v´

 

[wais]

Hi there!

To simplify the derivative of log(sqrt(1+log x)-sin x), we can follow these steps:

Step 1: Rewrite the expression as log(1+log x)^1/2 - log(sin x)

Step 2: Use the chain rule to find the derivative of log(1+log x)^1/2. The derivative of log(u) is 1/u * u'. In this case, u = (1+log x)^1/2 and u' = (1/2)(1+log x)^(-1/2) * (1/x). So, the derivative of log(1+log x)^1/2 is (1/2)(1+log x)^(-1/2) * (1/x).

Step 3: Use the chain rule again to find the derivative of log(sin x). The derivative of log(u) is 1/u * u'. In this case, u = sin x and u' = cos x. So, the derivative of log(sin x) is cos x / sin x.

Step 4: Combine the derivatives from steps 2 and 3 to get the final answer. The final answer is (1/2)(1+log x)^(-1/2) * (1/x) - cos x / sin x.

Hope this helps! Let me know if you have any further questions.