Find the derivative of the given function

[chwala]

Homework Statement

See attached( I want to attempt the problem using quotient and product rule).

Relevant Equations

Differentiation

Let's see how messy it gets...

$\dfrac{dy}{dx}=\dfrac{(1-10x)(\sqrt{x^2+2})5x^4 -(x^5)(-10)(\sqrt{x^2+2})-x^5(1-10x)\frac{1}{2}(x^2+2)^{-\frac{1}{2}}2x}{[(1-10x)(\sqrt{x^2+2})]^2}$

$\dfrac{dy}{dx}=\dfrac{5x^4(1-10x)(x^2+2)+(10x^5(x^2+2))-x^6(1-10x)}{\sqrt{x^2+2}}⋅\dfrac{1}{[(1-10x)(\sqrt{x^2+2})]^2}$$\dfrac{dy}{dx}=\dfrac{5x^4(1-10x)(x^2+2)+(10x^5(x^2+2))-x^6(1-10x)}{\sqrt{x^2+2}}⋅\dfrac{1}{[(1-10x)^2(\sqrt{x^2+2})^2]}$

$\dfrac{dy}{dx}=\dfrac{5x^4(1-10x)(x^2+2)+(10x^5(x^2+2))-x^6(1-10x)}{[(1-10x)^2(\sqrt{x^2+2})^3]}$

$\dfrac{dy}{dx}=\dfrac{5x^4}{(1-10x)\sqrt{x^2+2}}+\dfrac{10x^5}{[(1-10x)^2\sqrt{x^2+2}}-\dfrac{x^6}{[(1-10x)\sqrt{x^2+2})^3]}$

Factoring out $\dfrac{1}{(1-10x)\sqrt{x^2+2}}$ will give the desired result.

Bingo!!

 

[fresh_42]

This is definitely a case for WA:
https://www.wolframalpha.com/input?i=y=x^5/((1-10x)root(x^2+2))

 

[Mark44]

Let's see how messy it gets...

$\dfrac{dy}{dx}=\dfrac{(1-10x)(\sqrt{x^2+2})5x^4 -(x^5)(-10x)(\sqrt{x^2+2})-x^5(1-10x)\frac{1}{2}(x^2+2)^{-\frac{1}{2}}2x}{[(1-10x)(\sqrt{x^2+2})]^2}$

$\dfrac{dy}{dx}=\dfrac{5x^4(1-10x)(x^2+2)+(10x^6(x^2+2))-x^6(1-10x)}{\sqrt{x^2+2}}⋅\dfrac{1}{[(1-10x)(\sqrt{x^2+2})]^2}$

$\dfrac{dy}{dx}=\dfrac{5x^4(1-10x)(x^2+2)+(10x^6(x^2+2))-x^6(1-10x)}{\sqrt{x^2+2}}⋅\dfrac{1}{[(1-10x)^2(\sqrt{x^2+2})^2]}$

checking latex a minute

That's pretty messy. The logarithmic differentiation that was recommended seems to be a lot simpler.

 

[chwala]

That's pretty messy. The logarithmic differentiation that was recommended seems to be a lot simpler.

True...just a little exercise for the brain...

 

[pasmith]

I'd just like to note that, in the proposed solution by taking logs before differentiating, one should first simplify $\ln(x^5) = 5 \ln x$ and $\ln(\sqrt{x^2 + 2}) = \frac12\ln(x^2 + 2)$ before taking the derivative, thereby saving an application of the chain rule.