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I am to find the derivative of $y=\sin\left(t+\cos\left(\sqrt{t}\,\right)\right).$ This is what I have so far. \begin{align*} y&=\sin(t+\cos(\sqrt{t}\,)) \\ y'&=(\sin(t+\cos(\sqrt{t}\,)))' \cdot (t+\cos(\sqrt{t}\,))' \qquad \text{(chain rule)}\\ y'&= \cos(t+\cos(\sqrt{t}\,)) \cdot ( (t)' + (\cos(\sqrt{t}\,))' )\qquad\text{(sum rule)}\\ y'&= \cos(t+\cos(\sqrt{t}\,)) \cdot ( 1 -\sin(\sqrt{t}\,))\\ \end{align*} How do I proceed from here? At first I tried to FOIL, but that didn't work out:

\begin{align*} y'&= \cos(t + \cos(\sqrt{t}\,)) \cdot (-\sin(\sqrt{t}\,) + 1)\\ y'&= \cos(t + \cos(\sqrt{t}\,)) + \cos(t+\cos(\sqrt{t}\,)) \cdot (-\sin(\sqrt{t}\,) + 1)\\ y'&= \cos(t+\cos(\sqrt{t}\,)) + \cos(t+\cos(\sqrt{t}\,)) + (\cos(t+\cos(\sqrt{t}\,))) \cdot (-\sin(\sqrt{t}\,))\\ y'&= 2\cos(t+\cos(\sqrt{t}\,)) + (\cos(t+\cos(\sqrt{t}\,))) \cdot (-\sin(\sqrt{t}\,)) \end{align*}

Did I make a mistake somewhere? What am I missing here?

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    $\begingroup$ The derivative of $\cos\sqrt{t}$ is not $-\sin\sqrt{t}$ $\endgroup$ – A. Goodier May 29 '18 at 18:25
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You simply forgot that $\;(\cos\sqrt t)'=-\sin \sqrt t\,(\sqrt t)'=-\dfrac{\sin \sqrt t}{2\sqrt t}$, so the final result should be $$\Bigl(\sin\bigl(t+\cos\sqrt t\bigr)\Bigr)'=\cos\bigl(t+\cos\sqrt t\bigr)\biggl(1-\dfrac{\sin \sqrt t}{2\sqrt t}\biggr).$$

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    $\begingroup$ Can the maniac down-voter explain? Is my answer wrong? Did I not answer the O.P.'s question? $\endgroup$ – Bernard May 31 '18 at 19:58
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You have $$ \begin{split} \frac{d\left[\sin\left(t+\cos \sqrt{t}\right)\right]}{dt} &= \cos \left(t+\cos \sqrt{t}\right) \frac{d[t+\cos \sqrt{t}]}{dt}\\ &= \cos \left(t+\cos \sqrt{t}\right) \left(1 + \frac{d[\cos \sqrt{t}]}{dt}\right)\\ &= \cos \left(t+\cos \sqrt{t}\right) \left(1 - \sin\left( \sqrt{t} \right)\frac{d[\sqrt{t}]}{dt}\right)\\ &= \cos \left(t+\cos \sqrt{t}\right) \left(1 - \sin\left( \sqrt{t} \right)\frac{1}{2\sqrt{t}}\right)\\ &= \cos \left(t+\cos \sqrt{t}\right) \left(1 - \frac{\sin\left( \sqrt{t} \right)}{2\sqrt{t}}\right)\\ \end{split} $$

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Alternatively, if you want to show the steps of the chain rule, you can denote: $$y=\sin a, a=b^2+\cos b, b=\sqrt{t}.$$ Then: $$\frac{dy}{dt}=\frac{dy}{da}\cdot \frac{da}{db}\cdot \frac{db}{dt}=\cos a\cdot (2b-\sin b)\cdot \frac{1}{2\sqrt{t}}=\\ \cos (t+\cos \sqrt{t})\cdot (2\sqrt{t}-\sin \sqrt{t})\cdot \frac{1}{2\sqrt{t}}=\\ \cos (t+\cos \sqrt{t})\cdot \left(1-\sin \frac{\sqrt{t}}{2\sqrt{t}}\right).$$

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You wrote

$ y'=(\sin(t+\cos(\sqrt{t}\,)))' \cdot (t+\cos(\sqrt{t}\,))'$

The prime notation (also known as Lagrance notation) is generally understood to be the derivative with respect to whatever variable is given. So ($\sin(t+\cos(\sqrt{t}\,)))'$ is understood to be $\frac{d(\sin(t+\cos(\sqrt{t}\,)))}{dt}$. The correct statement, however, would be

$ y'=\frac{d(\sin(t+\cos(\sqrt{t}\,)))}{d(t+\cos(\sqrt{t})} \cdot \frac{d(t+\cos(\sqrt{t}\,))}{dt}$

In this case, you interpreted your notation $f'(u)$ as being $\frac{d f(u(t))}{d(u(t))}$, which is consistent with your nonstandard use, but not consistent with standard use, but later you interpreted $(\cos(\sqrt{t}\,))'$ as being $\frac{d(\cos(\sqrt{t}\,))}{dt}$, which is inconsistent with your previous usage of $f'(u)=\frac{d f(u(t))}{d(u(t))}$, albeit consistent with the standard meaning of $f'(u)$ as being $\frac{d f(u(t))}{dt}$. It is this inconsistent usage that led to your error.

Note that when given in the infinitesimal notation (also known as Leibniz notation), the quantity the derivative is with respect to is explicitly given, and the chain rule is a quite intuitive process of expanding out a "fraction" (note that it's not a fraction in the usual sense, as it is a fraction of infinitesimals). Thus the chain rule with Leibniz notation would look like this:

$\frac{d(\sin(t+\cos(\sqrt{t}\,)))}{dt} =$

$\frac{d(\sin(t+\cos(\sqrt{t}\,)))}{d(t+\cos(\sqrt{t})} \cdot \frac{d(t+\cos(\sqrt{t}\,))}{dt} =$

$\frac{d(\sin(t+\cos(\sqrt{t}\,)))}{d(t+\cos(\sqrt{t})} \cdot \left [\frac{dt+d\cos(\sqrt{t}\,))}{dt} \right ] =$

$\frac{d(\sin(t+\cos(\sqrt{t}\,)))}{d(t+\cos(\sqrt{t})} \cdot \left [\frac{dt}{dt}+\frac{d\cos(\sqrt{t}\,))}{dt} \right ] =$

$\frac{d(\sin(t+\cos(\sqrt{t}\,)))}{d(t+\cos(\sqrt{t})} \cdot \left [\frac{dt}{dt}+\frac{d\cos(\sqrt{t}\,))}{d (\sqrt t)} \cdot \frac{d ( \sqrt t)}{dt} \right ] $

Thus, it's harder to mess up the chain rule when using Leibnez notation. It's easy to get into the mindset of "the derivative of $sin$ is $cos$", but it's a lot safer if you write it as $d(sin(u))= cos(u)du$. Think of the symbol $d$ as representing something that you're trying to move to the right, and every function has a different rule for moving it to the right: $sin$ changes to $cos$, $x^n$ changes to $nx^{n-1}$, etc. So:

$d(\sin(t+\cos(\sqrt{t}\,)))=$

$\cos(t+\cos(\sqrt{t}\,)))d(t+\cos(\sqrt{t}))=$

$\cos(t+\cos(\sqrt{t}\,)))d(t^1+\cos(t^{1/2}))=$

$\cos(t+\cos(\sqrt{t}\,)))(1t^0dt-\sin(t^{1/2})d( t^{1/2}))=$

$\cos(t+\cos(\sqrt{t}\,)))(1t^0dt-\sin(t^{1/2})(1/2)t^{-1/2}dt)=$

$\cos(t+\cos(\sqrt{t}\,)))(dt-\frac{\sin(t^{1/2})}{2t^{1/2}}dt)=$

$\cos(t+\cos(\sqrt{t}\,)))(1-\frac{\sin(t^{1/2})}{2t^{1/2}})dt$

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$$y'= \cos(t+\cos√t) * ( 1 + \color{red}{-\sin√t)}$$ Note that you still have to use the chain rule here too

$$(\cos(\sqrt t))'=-\sin(\sqrt t) \times (\sqrt t)'$$ $$(\cos(\sqrt t))'=-\sin(\sqrt t) \times (t^{1/2})'$$ And $$(t^n)'=nt^{n-1}$$

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