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So I have the following:

$$ y = cos(a^3 + x^3) $$

This is what I got.

$$ y' = \cos(a^3 + x^3) \ ( -sin(a^3 + x^3) ) \ ( 3a^2 + 3x^2 ) $$

I'm not sure what to do after this? Would this be the final answer?

  • You started off with $y=\cos(a^2 + x^3)$ but when you took the derivative, it became $a^3$ in the $-\sin(...)$ part. – homegrown Feb 21 '14 at 05:33
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    First, the factor of $ \ \cos (a^2 + x^3) \ $ should not be there, since that is what you have differentiated. Second, in most of these expressions, letters from the beginning of the alphabet, such as $ \ a, b, c , ... \ $ represent unspecified constants, so their derivatives are zero. – colormegone Feb 21 '14 at 05:33
  • sorry the a and x should be $x^3 a^3$ –  Feb 21 '14 at 05:40
  • @RecklessReckoner thanks! the unspecified constants helped me see what i was doing wrong. –  Feb 21 '14 at 05:54
  • This is a notational convention that students innocently stumble over at some point. When I teach "Calc I", I make a point of explaining about "unspecified constants" in formulas and expressions, because otherwise I get asked what to do about the "other variable" $ \ a \ . $ – colormegone Feb 21 '14 at 05:58

3 Answers3

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$$d[\cos(a^3+x^3)]=-\sin(a^3+x^3)d(a^3+x^3)=-\sin(a^3+x^3)3x^2dx$$

assuming $a$ is a constant and $x$ is the variable. If you want to explicitly use the chain rule then let

$$u=a^3+x^3,\frac{du}{dx}=3x^2$$ $$y=\cos u,\frac{dy}{du}=-\sin u$$ $$\frac{dy}{dx}=\frac{du}{dx}\times\frac{dy}{du}=-3x^2\sin(a^3+x^3)$$

Mike
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What you wrote is not correct. Assuming $x$ is the variable, the correct answer is $y' = -\sin(a^2+x^3)3x^2$

Euler....IS_ALIVE
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There is a wrong cosine function, a mistake (probably misspelling) in the argument of sine and a mistake in the third multiplier.
Is $y'$ equal to $\dfrac{\text d y}{\text d x}$? And $a$ is a constant or function?

sas
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  • what do you mean by wrong cos function because this is the original problem here http://i.imgur.com/ZEm6Yda.png –  Feb 21 '14 at 05:49
  • sas was referring to the $ \ a^2 \ $ you had in the argument of the sine function, which you have since corrected. – colormegone Feb 21 '14 at 05:51
  • I am not clear, excuse me. You have a cosine function, yes, but when you differentiate it? $(\cos x)'$ isn't a $-\cos x \cdot \sin x$, it's just $-\sin x$. – sas Feb 21 '14 at 05:52