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So, this is the question:

Differentiating twice

But, how can it be $\cos^2x?$ Doesn't it mean to be just $cosx$? Do you think it is a typo?

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1 Answer 1

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This is just the product rule. Try this:
Let $f(x) = \cos x,\quad g(x)=\frac{\mathrm d y}{\mathrm dz}.$ Then $\frac{\mathrm d y}{\mathrm dx} = f(x)g(x)$ and so if we differentiate we get \begin{align} \frac{\mathrm d^2 y}{\mathrm dx^2} &= f'(x)g(x)+f(x)g'(x)\\ &=-\sin x\frac{\mathrm d y}{\mathrm dz} + \cos x \frac{\mathrm d }{\mathrm dx}\frac{\mathrm d y}{\mathrm dz}\\ &=-\sin x\frac{\mathrm d y}{\mathrm dz} + \cos x \frac{\mathrm d }{\mathrm dz}\frac{\mathrm d y}{\mathrm dx}\\ &=-\sin x\frac{\mathrm d y}{\mathrm dz} + \cos x \frac{\mathrm d }{\mathrm dz}\frac{\mathrm d y}{\mathrm dz}\cos x\\ &=-\sin x\frac{\mathrm d y}{\mathrm dz} + \frac{\mathrm d^2 y}{\mathrm dz^2}\cos^2 x\\ \end{align}

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  • $\begingroup$ You're missing a negative sign, should be ${-\sin{x}}$. I tried to edit it in, but it's only 4 characters and you need at least 10 :) $\endgroup$
    – Barry
    Jun 1, 2015 at 19:21
  • $\begingroup$ Yeah. I was thinking the same. ^^ $\endgroup$
    – Vaishnavi
    Jun 1, 2015 at 19:21
  • $\begingroup$ Opps, I had $-\sin x$ but must have dropped it by mistake $\endgroup$ Jun 1, 2015 at 19:23
  • $\begingroup$ Thanks a lot. @DanRobertson $\endgroup$
    – Vaishnavi
    Jun 1, 2015 at 19:24

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