basic differentiation of x = g(x) = sqrt((x^3 +11x -5)/7)

[pxy2d1]

I need to differentiate the following.

. . .x = g(x) = sqrt((x^3 +11x -5)/7)

At the moment, I can only differentiate very basic stuff, so would appreciate some help. Should I simplify to the following?

. . .((x^3 +11x -5)/7) ^1/2

I think I need the Chain Rule, but not sure how to do it.

Thank you!

 

[galactus]

Yes, use the chain rule.

\(\displaystyle \L\\\frac{d}{dx}\left[(\frac{x^{3}+11x-5}{7})^{\frac{1}{2}\right]\)

\(\displaystyle \L\\\underbrace{\frac{1}{2}(\frac{x^{3}+11x-5}{7})^{\frac{-1}{2}}}_{\text{outside}}\underbrace{(\frac{3x^{2}}{7}+\frac{11}{7})}_{\text{inside}}\)

\(\displaystyle \L\\\frac{\frac{3x^{2}}{7}+\frac{11}{7}}{2(\frac{x^{3}+11x-5}{7})^{\frac{1}{2}}}\)

Do some factoring:

\(\displaystyle \L\\\frac{\sqrt{7}(3x^{2}+11)}{14\sqrt{x^{3}+11x-5}}\)

 

[soroban]

Re: basic differentiation

Hello, pxy2d1!

Galactus did a great job!

Simplifying first is a splendid idea . . . if you're careful.

Differentiate: \(\displaystyle \L\,g(x) \:= \:\sqrt{\frac{x^3\,+\,11x\,-\,5}{7}}\)

We have: \(\displaystyle \L\,g(x) \:= \:\sqrt{\frac{x^3\,+\,11x\,-\,5}{7}} \;=\;\frac{\sqrt{x^3\,+\,11x\,-\,5}}{\sqrt{7}} \;=\;\frac{1}{\sqrt{7}}(x^3\,+\,11x\,-\,5)^{\frac{1}{2}}\;\;\) **

Chain Rule: \(\displaystyle \L\:g'(x)\;=\;\frac{1}{\sqrt{7}}\,\cdot\,\underbrace{\frac{1}{2}(x^3\,+\,11x\,-\,5)^{-\frac{1}{2}}}_{\text{outside}}\,\cdot\underbrace{(3x^2\,+\,11)}_{\text{inside}}\)

. . . . . . . \(\displaystyle \L g'(x)\;=\;\frac{3x^2\,+\,11}{2\sqrt{7}\sqrt{x^3\,+\,11x\,-\,5}} \;= \;\frac{\sqrt{7}(3x^2\,+\,11)}{14\sqrt{x^3\,+\,11x\,-\,5}}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

Whenever I see a constant in the numerator or denominator, I simplify first.

Example: \(\displaystyle \L\,f(x)\:=\:\left(\frac{x}{2}\right)^3\)

. . This is: \(\displaystyle \L\,f(x)\:=\:\frac{x^3}{8} \:=\:\frac{1}{8}x^3\)

. . And we don't need the Chain Rule: \(\displaystyle \L\,f'(x)\:=\;\frac{3}{8}x^2\)


Example: \(\displaystyle \L\:g(x)\;=\;\left(\frac{2}{x}\right)^3\)

. . This is: \(\displaystyle \L\,g(x)\;=\;\frac{8}{x^3}\;=\;8x^{-3}\)

. . Again, no Chain Rule: \(\displaystyle \L\,g'(x)\:=\:-24x^{-4} \:=\:-\frac{24}{x^4}\)