Derivative not correct for online hw submission, but is correct

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Homework Statement


Find the d/dx
(x^2+2x+5)^2


Homework Equations




Chain rule

The Attempt at a Solution



My answer:

2(x^2+2x+5)*(2x+2)*2
 
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Why did you apply the chain rule again to (2x+2)? You only need to do it once.

It should be:

##2(x^2 + 2x + 5)(2x + 2)##

Think of it this way: ##f(x) = x^2## and ##g(x) = x^2 + 2x + 5##.

Then, ##D(f \circ g)(x) = Df(g(x))D(g(x))##
 
Last edited:
You could also just multiply it out and use the power rule.
 
Torshi, your title is misleading. The derivative you show is not correct for online submission, AND it is not correct, as Karnage1993 points out.
 
Mark44 said:
Torshi, your title is misleading. The derivative you show is not correct for online submission, AND it is not correct, as Karnage1993 points out.

I didn't mean it to be misleading. My fault. The reason why I said that was because I was at my university math lab and even one of the math instructors said it was correct, but made a tad mistake until I mentioned it. I figured out the answer though.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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