Derivatives Practice

Geometrically, $f'(a)$ is the slope of the tangent line to the graph at $x=a$.

It also describes the function's instantaneous rate of change at that point.

Start with the definition:

Substitute $f(x)=x^2$:

Expand the numerator:

Simplify:

Now take the limit:

The derivative is written as

This is equivalent to $f'(x)$ when $y=f(x)$.

Use the power rule:

So

Differentiate term by term:

So the derivative is

Use the exponential rule:

With $a=2$,

Differentiate each term:

and

So the derivative is

Use the logarithmic derivative rule:

Use the product rule:

Let $f=x^2$ and $g=e^x$. Then $f'=2x$ and $g'=e^x$.

So

Use the quotient rule:

Let $f=x^2+1$ and $g=x$. Then $f'=2x$ and $g'=1$.

So

Simplify:

Use the chain rule. Differentiate the outside first and then multiply by the derivative of the inside:

So the derivative is

Treat the inside as the inner function. The derivative of $\sin u$ is $\cos u$, and the derivative of $x^3$ is $3x^2$.

So

Thus,

Use the chain rule with $u=2x^2-7x+4$.

The derivative of $\ln u$ is $\frac{1}{u}u'$.

Here,

So

Use the inverse trig rule and the chain rule:

Here $u=3x$, so $u'=3$.

Therefore,

Differentiate each term with respect to $x$:

Use the product rule on $xy$ and the chain rule on $y^2$:

Group the $\frac{dy}{dx}$ terms:

So

Velocity is the first derivative of position:

Acceleration is the derivative of velocity:

Now evaluate at $t=2$:

For $f(x)=\sqrt{x}$,

and

so

The linearization at $x=16$ is

Evaluate at $x=16.2$:

So

First find the critical numbers from $f'(x)=0$:

So the critical numbers are $x=-1$ and $x=2$.

Check the sign of $f'(x)$ on each interval:

• If $x<-1$, then both factors are negative, so $f'(x)>0$.

• If $-1<x<2$, then one factor is negative and one is positive, so $f'(x)<0$.

• If $x>2$, then both factors are positive, so $f'(x)>0$.

Therefore, $f$ is increasing on

and decreasing on

Use the area formula

Differentiate with respect to time:

Substitute $r=10$ and $\frac{dr}{dt}=0.5$:

So the area is changing at

Differentiate implicitly:

Solve for the derivative:

At $(3,5)$, the slope is

Use point-slope form:

Let the side lengths be $x$ and $y$.

The perimeter condition is

so

The area is

Differentiate:

Set the derivative equal to zero:

Then

So the rectangle with maximum area is a $10$ m by $10$ m square.

For a circle,

so

Substitute $r=10$ and $dr=0.05$:

So the area may be off by about

which is approximately $3.14\text{ cm}^2$.

Let

Then

Newton's method gives

With $x_0=1$:

So the next approximation is

First check continuity:

so $f$ is continuous at $0$.

Now check differentiability using one-sided slopes. For $x>0$, $f(x)=x$, so the right-hand derivative at $0$ is

For $x<0$, $f(x)=-x$, so the left-hand derivative at $0$ is

The one-sided derivatives are not equal, so $f'(0)$ does not exist.

Differentiate once:

So the critical points occur at

Differentiate again:

At $x=-1$,

so $x=-1$ is a local maximum.

At $x=1$,

so $x=1$ is a local minimum.

For an inflection point, check where concavity changes. Since

changes sign at $x=0$, there is an inflection point at $x=0$.

Let $x$ be the two equal sides perpendicular to the wall, and let $y$ be the side parallel to the wall.

The fencing constraint is

so

The area is

Differentiate:

Set the derivative equal to zero:

Then

So the maximum area occurs for dimensions $6$ m by $6$ m by $12$ m, with the $12$ m side against the wall.

Differentiate implicitly:

Use the product rule on $xy$:

Group the derivative terms:

So

At $(2,1)$,

Use point-slope form: