Computing $\displaystyle \small y'$ of the relation $\displaystyle \small x^2y+2x-y^2=x$ can be done through implicit differentiation:

$\displaystyle \small \small \frac{d}{dx}(x^2y+2x-y^2)=\frac{d}{dx}x$

$\displaystyle \small \small \implies 2xy+x^2y'+2-2yy'=1$

Now we isolate the $\displaystyle \small y'$:

$\displaystyle \small \small \small \iff x^2y'-2yy'=1-2-2xy=-2xy-1$

$\displaystyle \small \iff y'(x^2-2y)=-(2xy+1)$

$\displaystyle \small \small \implies y'=-\frac{2xy+1}{x^2-2y}$

This is a simple problem of integration. To find the formula for concentration from the formula of concentration rates, you simply take the integral of both sides of the rate equation with respect to time. 

$\displaystyle \int \frac{dC}{dt} dt=\int (k)dt$

Therefore, the concentration function is given by

$\displaystyle C=kt+C_0$, where $\displaystyle C_0$ is the initial concentration. 

Plugging in our values,

$\displaystyle C=3*10+40= 70M$

We will use the chain and power rules to differentiate both sides of this equation.

Power Rule: 

$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$

Chain Rule:

$\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$.

Applying the above rules to our function we find the following derivative.

$\displaystyle \frac{dy}{dt}=\frac{d}{dt}[3x^{4}-x+5]$

$\displaystyle \frac{dy}{dt}=12x^{3}\frac{dx}{dt}-1\frac{dx}{dt}$

at $\displaystyle t=2$, $\displaystyle \frac{dx}{dt}=6$ and $\displaystyle x=1$.

Therefore at $\displaystyle t=2$

$\displaystyle \frac{dy}{dt}=12(1)^{3}(6)-1(6)=72-6=66$

The form of log differentiation after first "logging" both sides, then taking the derivative is as follows:

$\displaystyle \frac{1}{y}y' = \frac{d}{dx}(lnf(x))$    which implies

$\displaystyle y' = f(x)\frac{d}{dx}(lnf(x))$

So:

$\displaystyle y'(e)=e^{elne-e}(ln^{2}e+2lne-\frac{e}{e})= 1(1+2-1)= 2$

Taking the derivative of the function:

$\displaystyle f'(x)= \frac{d}{dx}[3x^2 + x -1]$

$\displaystyle \frac{dy}{dx}= 6x + 1$

Evaluating at $\displaystyle x=3$:

$\displaystyle \frac{dy}{dx}|_{x=3} = 6(3) + 1$

$\displaystyle \frac{dy}{dx}|_{x=3} = 6(3) + 1=19$

Manipulating the equation:

$\displaystyle dy =19dx$

Allowing dx to be .01:

$\displaystyle dy=19*0.01$

$\displaystyle dy=0.19$

Which is our answer.

First, we take the derivative of the function:

$\displaystyle f'(x)=\frac{d}{dx} [\tan (\cos x)]$

$\displaystyle =\sec^2(\cos x)*(-\sin x)$

$\displaystyle =-\sin x \sec^2(\cos x)$

evaluate the derivative at $\displaystyle x= \frac{\pi}{4}$

$\displaystyle =-\sin (\frac{\pi}{4}) \sec^2(\cos (\frac{\pi}{4}))$

$\displaystyle =-\frac{\sqrt{2}}{2} \sec^2(\frac{\sqrt{2}}{2})$

Manipulating the equation by solving for dy:

$\displaystyle dy=-\frac{\sqrt{2}}{2} \sec^2(\frac{\sqrt{2}}{2})dx$

Assuming dx = 0.3

$\displaystyle dy=-\frac{\sqrt{2}}{2} \sec^2(\frac{\sqrt{2}}{2})*(0.3)$

$\displaystyle \approx -0.3669$

This is a related rate problem.  To find the rate of change of volume with respect to radius, we need to take the derivative of the volume of a sphere equation $\displaystyle (\frac{4}{3}\pi r^{3}).$

Then, we will plug in the relevant information.  The initial radius will be substituted in for $\displaystyle r$, and $\displaystyle dr=0.5in$, since that is the change from the initial to final radius of the balloon.

$\displaystyle dV=4\pi r^{2}dr.$

$\displaystyle dV=4\pi (5in)^{2}(0.5in)=50\pi in^{3}.$

To find the rate of change of a polynomial at a point, we must find the first derivative of the polynomial and evaluate the derivative at that point.

For this problem,

$\displaystyle y=4x^{3}-12x^{2}+6x+7$

the first derivative of this expression is

$\displaystyle \frac{dy}{dx}=12x^{2}-24x+6$

at $\displaystyle x=2$ the rate of change is 

$\displaystyle \frac{dy}{dx}=12*(2)^{2}-24(2)+6=48-48+6=6$