Derivatives - AP Calculus BC

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Question

What is the derivative of ?

Answer

Use Chain Rule. Identify and

,

Find the derivatives:

Use the formula:

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Question

What is the derivative of $y=\frac {2x^2+1}{x^3-2x}$

Answer

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Question

Find the derivative of $\small lnx^{lnx}$ .

Answer

To solve this derivative, we need to use logarithmic differentiation. This allows us to use the logarithm rule $\small lna^{b}=blna$ to solve an easier derivative.

Let $\small y=lnx^{lnx}$ .

Now we'll take the natural log of both sides to get

$\small lny=ln(lnx^{lnx})=lnxln(lnx)$ .

Now we can use implicit differentiation to solve for $\small y'$ .

The derivative of $\small lny$ is $\small y'/y$ , and the derivative of $\small \small \small lnx(ln(lnx))$ can be found using the product rule, which states

$\small \frac{\mathrm{d} }{\mathrm{d} x}(u\cdot v)=u'v+uv'$ where $\small u$ and $\small v$ are functions of $\small x$ .

Letting $\small u=lnx$ and $\small v=ln(lnx)$

(which means $\small u'=1/x$ and $\small v'=1/(xlnx)$ ) we get our derivative to be $\small ln(lnx)/x+1/x$ .

Now we have $\small y'/y=ln(lnx)/x+1/x$ , but $\small y=lnx^{lnx}$ , so subbing that in we get

$\small \small y'/(lnx^{lnx})=ln(lnx)/x+1/x$ .

Multiplying both sides by $\small lnx^{lnx}$ , we get

$\small y'=lnx^{lnx}(ln(lnx)/x+1/x)$ .

That is our derivative.

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Question

The position of a particle is given by $s(t)=e^{t^2}+t^2$ . Find the acceleration of the particle when .

Answer

The acceleration of a particle is given by the second derivative of the position function. We are given the position function as

$s(t)=e^{t^2}+t^2$ .

The first derivative (the velocity) is given as

$\frac{d(e^{t^2}+t^2)}{dt}=2te^{x^2}+2t$ .

The second derivative (the acceleration) is the derivative of the velocity function. This is given as

$\frac{d(2te^{t^2}+2t)}{dt}=2e^{t^2}+4t^2e^{t^2}+2$ .

Evaluating this at gives us the answer. Doing this we get

$a(1)=2e^{1^2}+41^2e^{1^2}+2=6e+2=2(3e+1)$ .

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Question

The position of a particle is given by $s(t)=\frac{2t^2+1}{t+1}$ . Find the velocity at .

Answer

The velocity is given as the derivative of the position function, or

$v(t)=\frac{d(s(t))}{dt}$ .

We can use the quotient rule to find the derivative of the position function and then evaluate that at . The quotient rule states that

$\left[\frac{f(x)}{g(x)}\right]'=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$ .

In this case, $f(t)=2t^2+1\Rightarrow f'(t)=4t$ and $g(t)=t+1\Rightarrow g'(t)=1$ .

We can now substitute these values in to get

$\left[\frac{f(t)}{g(t)}\right]'=\frac{(t+1)(4t)-(2t^2+1)(1)}{(t+1)^2}=\frac{2t^2+4t-1}{(t+1)^2}$ .

Evalusting this at gives us $\frac{22^2+42-1}{(2+1)^2}=\frac{16-1}{3^2}=\frac{15}{9}$ .

So the answer is $\frac{15}{9}$ .

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Question

$f (x) = 7x^{2} + 6x - 9$

Give .

Answer

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

$f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$f ' (x) = 14\cdot x^{1} + 6$

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Question

Differentiate $x^{999}$ .

Answer

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

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Question

$f(x)= 8x^{3}-7x^{2}+11$

Give .

Answer

First, find the derivative of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate to get .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

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Question

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

Give .

Answer

First, find the derivative of $f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$ .

Recall that $\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0.

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate to get .

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$f''(x) =10.8x^{2} + 4.2x$

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Question

The position of an object is given by the following equation:

$s(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t+27$

Determine the equation for the velocity of the object.

Answer

Velocity is the derivative of position, so in order to find the equation for the velocity of an object, all we must do is take the derivative of the equation for its position:

$s(t)=\frac{2}{3}t^3+\frac{5}{2}t^2-13t+27$

We will use the power rule to get the derivative.

$f(x)=ax^n \rightarrow f'(x)=a \cdot n x^{n-1}$

Therefore we get,

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Question

The position of an object is described by the following equation:

$s(t)=\frac{5}{4}t^3-3t^2+\frac{1}{2}t-6$

Find the acceleration of the object at second.

Answer

Acceleration is the second derivative of position, so we must first find the second derivative of the equation for position:

$s(t)=\frac{5}{4}t^3-3t^2+\frac{1}{2}t-6$

$v(t)=s'(t)=\frac{15}{4}t^2-6t+\frac{1}{2}$

$a(t)=v'(t)=s''(t)=\frac{15}{2}t-6$

Now we can plug in t=1 to find the acceleration of the object after 1 second:

$a(1)=\frac{15}{2}(1)-6=\frac{3}{2}$

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Question

The position of a particle is represented by $f(x) = e^{x}+3x^2$ . What is the velocity at ?

Answer

Differentiate the position equation, to get the velocity equation

$f(x) = e^{x}+3x^2$

$v(x) = e^x +3\cdot2 x^{2-1}=e^{x} + 6x$

Now we plug 4 into the equation to find the velocity

$v(4) = e^{(4)}+6(4)$

is approximately equal to 2.72. Therefore

$e^{(4)} + 6(4) = 2.72^{(4)} +24 \approx 55+24= 79$

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Question

What is the velocity function when the position function is given by

.

Answer

To find the velocity function, we need to find the derivative of the position function.

So lets take the derivative of with respect to .

The derivative of is because of Power Rule:

$f(x)=a^n \rightarrow \ f'(x)=na^{n-1}$

$p(t)=t^2 \rightarrow p'(t)=2t^{2-1}=2t$

The derivative of is due to Power Rule

$p(t)=t \rightarrow p'(t)=1t^{1-1}=1t^0=1(1)=1$

So...

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Question

If models the distance of a projectile as a function of time, find the acceleration of the projectile at .

$h(t)=\frac{4x^3}{3}+\frac{4.5x^2}{3}+3x-7$

Answer

We are given a function dealing with distance and asked to find an acceleration. recall that velocity is the first derivative of position and acceleration is the derivative of velocity. Find the second derivative of h(t) and evaluate at t=6.

$h(t)=\frac{4t^3}{3}+\frac{4.5t^2}{3}+3t-7$

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Question

The position at a certain point is given by:

$f(x)=\frac{3}{5}x^{5} - 6x + 5$

What is the acceleration at ?

Answer

In order to find the acceleration of a given point, you must first find the derivative of the position function which gives you the velocity function:

$f'(x)=3x^{4}-6$

Then, you differentiate the velocity function to get the acceleration function:

Then, find when .

The answer is:

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Question

Function gives the velocity of a particle as a function of time.

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Find the equation that models that particle's acceleration over time.

Answer

Recall that velocity is the first derivative of position, and acceleration is the second derivative of position. We begin with velocity, so we need to integrate to find position and derive to find acceleration.

To derive a polynomial, simply decrease each exponent by one and bring the original number down in front to multiply.

So this

$\small \small v(t)=t^5+3t^4-\frac{7t^2}{3}$

Becomes:

$\small \small \small \small \small v'(t)=5t^4+12t^3-\frac{14t}{3}$

So our acceleration is given by

$\small \small \small \small \small \small a(t)=5t^4+12t^3-\frac{14t}{3}$

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Question

Consider the following position function:

Find the acceleration after seconds of a particle whose position is given by .

Answer

Recall that acceleration is the second derivative of position, so we need p''(7).

Taking the first derivative we get:

Taking the second derivative and plugging in 7 we get:

$p''(7)=72\cdot7^2-30\cdot7=3318$

So our acceleration after 7 seconds is

$3318 \frac{m}{s^2}$ .

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Question

Given the vector position:

$\vec{r}=(tcos(t),t, tsin(t)).$

Find the expression of the velocity.

Answer

All we need to do to find the components of the velocity is to differentiate the components of the position vector with respect to time.

We have :

Collecting the components we have :

$\vec{v}=(cos(t)-tsin(t),1,sin(t)+tcos(t))$

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Question

A car is driving north on a highway at a constant velocity of mph. What is the acceleration after an hour?

Answer

If a car is travelling north at constant velocity 60 mph, it's possible to write a velocity function for this vehicle, where is time in hours.

To find the acceleration, take the derivative of the velocity function.

The acceleration after an hour, or any time , is zero.

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Question

The displacement of an object at time is defined by the equation $f(t) = \frac{100}{t}$ . What is the acceleration equation for this object?

Answer

The acceleration equation is the second derivative of the displacement equation.

$f(t) = \frac{100}{t} = 100t^{-1}$

Therefore the first derivative is equal to

$f'(t) = (-1)(100)t^{-1-1} = -100t^{-2}$

Differentiating a second time gives

$f''(t) = (-2)(-100)t^{-2-1} = 200t^{-3} = \frac{200}{t^3}$

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