The rate of change of a function is the amount it changes over a given amount of time.

In mathematical terms, this can be written as

\(\displaystyle r=\frac{\Delta y}{\Delta x}\)

we plug in our values:

\(\displaystyle \frac{5^2+5+1-0^2-0-1}{5-0} = \frac{30}{5}=6\)

Note that this is only an average because quadratic functions change at different rates depending on where you are in the function's domain.

Note that:

\(\displaystyle f(1)= ln(e-sin(0))= lne =1\)

Therefore, for values relatively close  to 1, we can use the formula for dy (the differential form of the derivative) to estimate f at close values.

\(\displaystyle dy= \frac{e^{x}+\pi cos(\pi -\pi x)}{e^{x}-sin(\pi -\pi x)}dx\)  From log derivative with chain rule.

Since \(\displaystyle \frac{3}{2}\) lies \(\displaystyle \frac{1}{2}\) to the right of \(\displaystyle 1\), \(\displaystyle dx = \frac{1}{2}\) and \(\displaystyle x=1\) for the estimation, so:

\(\displaystyle dy = \frac{e^{3/2}+\pi cos(-\pi )}{e^{3/2}} = 1-\frac{\pi }{e^{3/2}}\)

So then:

\(\displaystyle f(3/2) \approx1+1-\frac{\pi }{e^{3/2}} = 2-\frac{\pi }{e^{3/2}}\)

First, we need to rearrange the given to match the approximation formula.  \(\displaystyle 0.985 = 1-0.015.\)  Therefore, \(\displaystyle (0.985)^{50}=(1-0.015)^{50}=1-50(0.015)=0.25.\textup{}\)

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

\(\displaystyle V=s^3\)

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

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

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

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

\(\displaystyle \phi=3s^2\)

\(\displaystyle \phi =3(\frac{7}{30})^2=\frac{49}{300}\)

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

\(\displaystyle V=s^3\)

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

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

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

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

\(\displaystyle \phi=3s^2\)

\(\displaystyle \phi =3(\frac{13}{15})^2=\frac{169}{75}\)

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

\(\displaystyle V=s^3\)

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

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

Now, knowing the length of the sides, simply divide to find the ratio between the rate of change of the volume and sides:

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

\(\displaystyle \phi=3s^2\)

\(\displaystyle \phi =3(\frac{1}{30})^2=\frac{1}{300}\)

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:

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