How to find the derivate of this function ?

[Yankel]

Hello all

I am interested in finding the derivative of the following function:

\[f(x)=)\sqrt[m+n]{(1-x)^{m}\cdot (1+x)^{n}}\]Can you please assist ? I tried transforming the root into 1 power of 1/m+n, but got stuck quite afterwards.

thank you

 

[MarkFL]

Yes, the first thing I would do is convert from radical notation to rational power:

$$f(x)=\left((1-x)^m(1+x)^n\right)^{\frac{1}{m+n}}$$

Next...to proceed with the differentiation w.r.t $x$, we need to apply the power rule, and the chain rule, which will involve the product/power rules.

$$f'(x)=\frac{1}{m+n}\left((1-x)^m(1+x)^n\right)^{\frac{1}{m+n}-1}\frac{d}{dx}\left(1-x)^m(1+x)^n\right)$$

So, let's turn our attention to:

$$\frac{d}{dx}\left(1-x)^m(1+x)^n\right)$$

What do you get when you apply the product/power/chain rules?

 

[Yankel]

I get this :

\[n\cdot \left ( 1+x \right )^{n-1}\cdot \left ( 1-x \right )^{m}-m\cdot \left ( 1+x \right )^{n}\cdot \left ( 1-x \right )^{m-1}\]inner derivative.

 

[MarkFL]

Okay, so what we have now is:

$$f'(x)=\frac{1}{m+n}\left((1-x)^m(1+x)^n\right)^{\frac{1}{m+n}-1}\left(n(1+x)^{n-1}(1-x)^m-m(1+x)^n(1-x)^{m-1}\right)$$

And with some factoring, we have:

$$f'(x)=\frac{1}{m+n}\left((1-x)^m(1+x)^n\right)^{\frac{1}{m+n}-1}(1+x)^{n-1}(1-x)^{m-1}\left(n(1-x)-m(1+x)\right)$$

Using the rules of exponents, we can then write:

$$f'(x)=\frac{1}{m+n}(1+x)^{-\frac{m}{m+n}}(1-x)^{-\frac{n}{m+n}}\left(n-m-(m+n)x\right)$$