Find the derivative of the given function

chwala
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Homework Statement
See attached( I want to attempt the problem using quotient and product rule).
Relevant Equations
Differentiation
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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!! :cool:
 
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chwala said:
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.
 
Mark44 said:
That's pretty messy. The logarithmic differentiation that was recommended seems to be a lot simpler.
True...just a little exercise for the brain... :cool:
 
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.
 
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