Begin by writing the equations for a cube's dimensions. Namely its volume and surface area in terms of the length of its sides:

\(\displaystyle V=s^3\)

\(\displaystyle A=6s^2\)

The rates of change of these can be found by taking the derivative of each side of the equations with respect to time:

\(\displaystyle \frac{dV}{dt}=3s^2\frac{ds}{dt}\)

\(\displaystyle \frac{dA}{dt}=12s\frac{ds}{dt}\)

Now, solve for the length of the side of the cube to satisfy the problem condition,

\(\displaystyle 3s^2\frac{ds}{dt}=12s\frac{ds}{dt}\)

\(\displaystyle s=4\)

Begin by writing the expression for the volume of a rectangular prism:

\(\displaystyle V=whl\)

The rate of change of the volume can be found by taking the derivative of the equation with respect to time:

\(\displaystyle \frac{dV}{dt}=wh\frac{dl}{dt}+wl\frac{dh}{dt}+hl\frac{dw}{dt}\)

Now, we're given some information:

The width of a rectangular prism increases three times as fast as its length and twice as fast as its height:\(\displaystyle \frac{dw}{dt}=3\frac{dl}{dt},\frac{dw}{dt}=2\frac{dh}{dt}\)

The length, height, and width are equal: \(\displaystyle w=l=h\)

Using this, rewrite the volume equation in terms of width:

\(\displaystyle \frac{dV}{dt}=w^2(\frac{1}{3}\frac{dw}{dt})+w^2(\frac{1}{2}\frac{dw}{dt})+w^2\frac{dw}{dt}\)

\(\displaystyle \frac{dV}{dt}=\frac{11}{6}w^2\frac{dw}{dt}\)

The rate of change of the volume is \(\displaystyle \frac{11}{6}w^2\) times the rate of change of the rate of change of the width.

Using \(\displaystyle f'(x)=\lim_{d \rightarrow 0} \frac{f(x+d)-f(x)}{d}\) ,

We have that:

\(\displaystyle f'(x)=\lim_{d \rightarrow 0} \frac{[(x+d)^3-6]-[x^3-6]}{d}\) Which will give us the general derrivative of the function. Expanding, we have:

\(\displaystyle =\lim_{d \rightarrow 0} \frac{x^3+3x^2d+3xd^2+d^3-6-x^3+6}{d}\)

Simplifying:

\(\displaystyle =\lim_{d \rightarrow 0} \frac{3x^2d+3xd^2+d^3}{d}\)

Factoring out a d:

\(\displaystyle =\lim_{d \rightarrow 0} \frac{d(3x^2+3xd+d^2)}{d}\)

canceling out the \(\displaystyle d\)'s

\(\displaystyle =\lim_{d \rightarrow 0}3x^2+3xd+d^2\)

Applying the limit, we have:

\(\displaystyle =\lim_{d \rightarrow 0}3x^2+3x(0)+0 =3x^2\)

Now, to find the instantaneous rate of change at \(\displaystyle f(-2)\) we evaluate the derrivative at \(\displaystyle -2\):

\(\displaystyle 3(-2)^2=3\cdot4=12\), which is our answer.

We'll try to find the derivative:

\(\displaystyle f'(x)=\frac{d}{dx}[\frac{1}{3} \ln x + \sin x -\cos x]\):

This is a sum of three distinct terms, so we can apply the derivative to each term seperately:

\(\displaystyle =\frac{d}{dx}[\frac{1}{3} \ln x] + \frac{d}{dx}[\sin x ]-\frac{d}{dx}[\cos x]\)

We apply our knowledge of derivatives of the various functions:

\(\displaystyle \frac{d}{dx} \ln x = \frac{1}{x}\) ; \(\displaystyle \frac{d}{dx} \sin x = \cos x\) ; \(\displaystyle \frac{d}{dx} \cos x = -\sin x\)

So, we have:

\(\displaystyle =\frac{1}{3} \cdot \frac{1}{x} + \cos x - (-\sin x)\)

Simplifying:

\(\displaystyle = \frac{1}{3x} + \cos x +\sin x\).

Evaluating this at \(\displaystyle x=\frac{\pi}{3}\), we get:

\(\displaystyle = \frac{1}{3 (\frac{\pi}{3})} + \cos (\frac{\pi}{3}) +\sin (\frac{\pi}{3})\)

\(\displaystyle = \frac{3}{3\pi} + \frac{1}{2} +\frac{\sqrt{3}}{2}\)

Equivalently:

\(\displaystyle = \frac{1}{\pi} +\frac{1+\sqrt{3}}{2}\)  which is our answer.

We find the derivative with respect to \(\displaystyle y\).

\(\displaystyle f'(y) = \frac{d}{dx}[\frac{1}{2} \tan y +\frac{7}{12} 3^y - \log_5{y}]\)

As a sum of different functions, the derrivative can be applied to each of the different terms separately. 

\(\displaystyle f'(y) = \frac{d}{dx}[\frac{1}{2} \tan y] +\frac{d}{dx}[\frac{7}{12} 3^y] - \frac{d}{dx}[\log_5{y}]\)

Now, to simplify, we can move the coeficient out of the braces.

\(\displaystyle = \frac{1}{2} \frac{d}{dx}[\tan y] +\frac{7}{12}\frac{d}{dx}[ 3^y] - \frac{d}{dx}[\log_5{y}]\)

We use the facts

\(\displaystyle \frac{d}{dy} \tan y =\sec^2 y\) ; \(\displaystyle \frac{d}{dy} a^y= \ln a \cdot a^y\) and \(\displaystyle \frac{d}{dy} \log_a y= \frac{1}{(\ln a )y}\)

to differentiate the function:

\(\displaystyle = \frac{1}{2} \sec^2 y +\frac{7}{12}\ln3\cdot3^y - \frac{1}{(\ln5) y}\) giving us the answer.

Taking the derrivative

\(\displaystyle f'(z)= \frac{d}{dx}[\frac{\sin^{-1}z}{5}-\frac{\cos^{-1}z}{3}+\frac{\tan^{-1}z}{\pi}]\)

We can differentiate each term separately.

\(\displaystyle =\frac{d}{dx}[\frac{\sin^{-1}z}{5}]-\frac{d}{dx}[\frac{\cos^{-1}z}{3}]+\frac{d}{dx}[\frac{\tan^{-1}z}{\pi}]\)

Factoring out the coeficients

\(\displaystyle =\frac{1}{5}\frac{d}{dx}[\sin^{-1}z]-\frac{1}{3}\frac{d}{dx}[\cos^{-1}z]+\frac{1}{\pi}\frac{d}{dx}[\tan^{-1}z]\)

We use the normal derivative rules:

\(\displaystyle =\frac{1}{5}\cdot \frac{1}{\sqrt{1-z^2}}-\frac{1}{3}\cdot (-\frac{1}{\sqrt{1-z^2}})+\frac{1}{\pi}\cdot \frac{1}{1+z^2}\)

Simplifying:

\(\displaystyle =(\frac{1}{5}+\frac{1}{3})\cdot \frac{1}{\sqrt{1-z^2}}+\frac{1}{\pi}\cdot \frac{1}{1+z^2}\)

\(\displaystyle =\frac{8}{15}\cdot \frac{1}{\sqrt{1-z^2}}+\frac{1}{\pi}\cdot \frac{1}{1+z^2}\)

Evaluating at \(\displaystyle x=0\)

\(\displaystyle f'(0)=\frac{8}{15}\cdot \frac{1}{\sqrt{1-(0)}}+\frac{1}{\pi}\cdot \frac{1}{1+(0)}\)

Simplifying

\(\displaystyle =\frac{8}{15}+\frac{1}{\pi}\)

Which is our answer.

We take the derivative of the function:

\(\displaystyle f'(x)=\frac{d}{dx}[\frac{1}{3} x^{-3} \arctan x]\)

We notice that this function is the product of two functions, so we use the product rule:

\(\displaystyle =\frac{d}{dx}[\frac{1}{3}x^{-3}]\arctan x+\frac{1}{3}x^{-3}\frac{d}{dx}[\arctan x]\)

\(\displaystyle =\frac{1}{3}\cdot (-3)x^{-4}\arctan x+\frac{1}{3}x^{-3}\frac{1}{1+x^2}\)

Simplifying, we get:

\(\displaystyle =-x^{-4}\arctan x+\frac{1}{3x^3(1+x^2)}\).

Which is equivalent to:

\(\displaystyle =-\frac{\arctan x}{x^4} +\frac{1}{3x^3(1+x^2)}\), our answer

taking the derivative

\(\displaystyle f'(x)=\frac{d}{dx}[\frac{\sin\alpha \cdot \log_3 y}{4^y}]\)

we notice we can factor out a coeficient:

\(\displaystyle =\sin\alpha \cdot\frac{d}{dx}[\frac{\log_3 y}{4^y}]\)

 then, we apply the quotient rule:

\(\displaystyle =\sin\alpha \cdot(\frac{\frac{d}{dx}[\log_3 y]\cdot 4^y- \log_3 y\cdot \frac{d}{dx}[4y]}{(4^y)^2})\)

\(\displaystyle =\sin\alpha \cdot (\frac{\frac{1}{y\ln3}\cdot 4^y- \log_3 y\cdot 4^y\ln 4}{4^{2y}})\)

taking out a common factor:

\(\displaystyle =\sin\alpha \cdot 4^y(\frac{\frac{1}{y\ln3}-\ln 4 \log_3 y}{4^{2y}})\)

and simplifying,

\(\displaystyle =\sin\alpha \cdot 4^{-y}(\frac{1}{y\ln3}-\ln 4 \log_3 y)\)

we have our answer.

Begin by writing the volume equations for a sphere and a cube:

\(\displaystyle V_s=\frac{4}{3}\pi r^3\)

\(\displaystyle V_c = s^3\)

The rate of change of these volumes can be found by taking the derivatives of these equation with respect to time:

\(\displaystyle \frac{dV_s}{dt}=4\pi r^2 \frac{dr}{dt}\)

\(\displaystyle \frac{dV_c}{dt}=3s^2 \frac{ds}{dt}\)

We're given the relationship between the rate of expansion of the pertinent lengths for these shapes, namely \(\displaystyle \frac{dr}{dt}=\frac{ds}{dt}\).

To find the necessary ratios of lengths for the volumes to have equal rates of expansion, set the volume rate expressions equal to each other:

\(\displaystyle \frac{dV_s}{dt}=\frac{dV_c}{dt}\)

\(\displaystyle 4\pi r^2 \frac{dr}{dt}=3s^2 \frac{ds}{dt}\)

\(\displaystyle 4\pi r^2 =3s^2\)

\(\displaystyle \frac{r^2}{s^2}=\frac{3}{4\pi}\)

\(\displaystyle \frac{r}{s}=\sqrt{\frac{3}{4\pi}}\)

When the velocity is \(\displaystyle 0\), that means \(\displaystyle p'(t)=0\). That is, \(\displaystyle 6t-12=0\). So \(\displaystyle t=2\).