MHB Differentiation with square roots

[Yankel]

Hello all,

I was trying to find derivatives of two functions containing square roots. I got answers which I believe should be correct, however, the answers in the book differ significantly. The first answer of mine was checked in MAPLE and found correct. My guess that the author made some algebraic manipulations but I can't seem to track it down and get to the same result. Can you kindly take a look ?

The functions are:

\[f(x)=(\sqrt{x}+\frac{1}{\sqrt{x}})^{10}\]

\[g(x)=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{x-1}}\]

My answers:

\[f'(x)=5\cdot (\sqrt{x}+\frac{1}{\sqrt{x}})^{9}\cdot (\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x^{3}}})\]

\[g'(x)=\frac{\frac{1}{2\sqrt{x}}\cdot (\sqrt{x}-\sqrt{x-1})-\sqrt{x}(\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x-1}})}{(\sqrt{x}-\sqrt{x-1})^{2}}\]Books answers:

\[f'(x)=\frac{5\cdot (x+1)^{9}\cdot (x-1)}{x^{6}}\]

\[g'(x)=1+\frac{2x-1}{2\sqrt{x^{2}-x}}\]Thank you !

 

[MarkFL]

For the first function:

$$f(x)=\left(x^{\Large\frac{1}{2}}+x^{-\Large\frac{1}{2}}\right)^{10}$$

Using the power and chain rules, I get:

$$f'(x)=10\left(x^{\Large\frac{1}{2}}+x^{-\Large\frac{1}{2}}\right)^{9}\left(\frac{1}{2}x^{-\Large\frac{1}{2}}-\frac{1}{2}x^{-\Large\frac{3}{2}}\right)$$

Factor:

$$f'(x)=5x^{-\Large\frac{3}{2}}(x-1)\left(x^{\Large\frac{1}{2}}+x^{-\Large\frac{1}{2}}\right)^{9}$$

Now, we can write:

$$f'(x)=5x^{-\Large\frac{3}{2}}(x-1)\left(\frac{x+1}{\sqrt{x}}\right)^{9}$$

$$f'(x)=5x^{-6}(x-1)(x+1)^9=\frac{5(x-1)(x+1)^9}{x^6}$$

 

[MarkFL]

For the second function, let's rationalize the denominator before differentiating:

$$g(x)=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{x-1}}\cdot\frac{\sqrt{x}+\sqrt{x-1}}{\sqrt{x}+\sqrt{x-1}}=x+\sqrt{x^2-x}$$

So, then:

$$g'(x)=1+\frac{2x-1}{2\sqrt{x^2-x}}$$

 

[I like Serena]

\[f(x)=(\sqrt{x}+\frac{1}{\sqrt{x}})^{10}\]

For the first function it also makes sense to rationalize first:
\[f(x)=\left(\frac{x}{\sqrt{x}}+\frac{1}{\sqrt{x}}\right)^{10}
=\left(\frac{x+1}{\sqrt{x}}\right)^{10}
=\frac{(x+1)^{10}}{x^{5}}\]
making it a bit easier to find the derivative.

We can do the same rationalization afterwards giving the same result.