AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

Example Question #254 : Derivatives

\(\displaystyle y+ \cos y = x^{2}\).

Which of the following expressions is equal to \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}\) ?

Possible Answers:

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \sin y }\)

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \tan y }\)

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 + \tan y }\)

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{ \sin y -1 }\)

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \cos y }\)

Correct answer:

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \sin y }\)

Explanation:

Differentiate both sides with respect to \(\displaystyle x\):

\(\displaystyle y+ \cos y = x^{2}\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(y+ \cos y )= \frac{\mathrm{d} }{\mathrm{d} x}(x^{2})\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(y+ \cos y )=2x\)

By the sum rule:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(y )+ \frac{\mathrm{d} }{\mathrm{d} x}( \cos y ) =2x\)

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} + \frac{\mathrm{d} }{\mathrm{d} x}( \cos y ) =2x\)

By the chain rule:

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} + \frac{\mathrm{d} y}{\mathrm{d} x} \frac{\mathrm{d} }{\mathrm{d} y}( \cos y ) =2x\)

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} + \frac{\mathrm{d} y}{\mathrm{d} x} \cdot (-\sin y) =2x\)

Applying some algebra:

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x} \left (1- \sin y \right ) =2x\)

\(\displaystyle \frac{\frac{\mathrm{d}y }{\mathrm{d} x} \left (1- \sin y \right )}{1 - \sin y } =\frac{2x}{1 - \sin y }\)

\(\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \sin y }\)

Example Question #471 : Ap Calculus Ab

\(\displaystyle y^{2}- 4y + 8 = x\)

Which of the following is equal to \(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}\) ?

Possible Answers:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{y}{ 2y - 4 }\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{x}{ y - 4 }\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{1}{ y - 4 }\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{1}{ 2y - 4 }\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{x}{ 2y - 4 }\)

Correct answer:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{1}{ 2y - 4 }\)

Explanation:

Differentiate both sides with respect to \(\displaystyle x\):

\(\displaystyle y^{2}- 4y + 8 = x\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} (y^{2}- 4y + 8 )=\frac{\mathrm{d} }{\mathrm{d} x} (x)\)

Apply the sum, difference, and constant multiple rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} (y^{2})- \frac{\mathrm{d} }{\mathrm{d} x} (4y )+\frac{\mathrm{d} }{\mathrm{d} x} ( 8 )=\frac{\mathrm{d} }{\mathrm{d} x} (x)\)

In the first term, apply the chain rule; in the second, apply the constant multiple rule:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} y} (y^{2})- 4 \frac{\mathrm{d} y}{\mathrm{d} x} +0=1\)

Apply the power rule:

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} \cdot 2y - 4 \frac{\mathrm{d} y}{\mathrm{d} x} =1\)

Now apply some algebra:

\(\displaystyle ( 2y - 4 )\frac{\mathrm{d} y}{\mathrm{d} x} =1\)

\(\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x} =\frac{1}{ 2y - 4 }\)

Example Question #164 : Computation Of The Derivative

We have three functions, \(\displaystyle g(x)=cos(x), h(x)=sin(x), k(x)=x^2\)

Find the derivative of \(\displaystyle f(x)\) 

Given that \(\displaystyle f(x)=g(h(k(x)))\)

Possible Answers:

\(\displaystyle -sin(sin(x^2))*cos(x^2)*2x\)

\(\displaystyle -sin(cos(2x))\)

\(\displaystyle sin(sin(x^2))*cos(x^2)\)

\(\displaystyle sin(sin(x^2))*cos(x^2)*2x\)

\(\displaystyle -sin(sin(x^2))*cos(2x)\)

Correct answer:

\(\displaystyle -sin(sin(x^2))*cos(x^2)*2x\)

Explanation:

So now this is a three layer chain rule differentiation. The more functions combine to form the composite function the harder it will be to keep track of the derivative. I find it helpful to lay out each equation and each derivative, so:

\(\displaystyle g(x)=cos(x), g'(x)=-sin(x)\)

\(\displaystyle h(x)=sin(x), h'(x)=cos(x)\)

\(\displaystyle k(x)=x^2, k'(x)=2x\)

Then a three layer chain rule is just the same as a two layer, except... there's one more layer!

\(\displaystyle f'(x)=g'(h(k(x)))*h'(k(x))*k'(x)\)

It is still the outermost layer evaluated at the inner layers, and then move another layer in and repeat

Example Question #257 : Derivatives

Find \(\displaystyle \frac{dy}{dx}\) at \(\displaystyle x=4\) with the equation \(\displaystyle x^2+y^2=25\) 

Possible Answers:

\(\displaystyle \frac{x}{y}\)

\(\displaystyle \frac{-4}{3}\)

\(\displaystyle 8+y\)

\(\displaystyle \frac{-4}{y}\)

\(\displaystyle \frac{-4}{3} or \frac{4}{3}\)

Correct answer:

\(\displaystyle \frac{-4}{3} or \frac{4}{3}\)

Explanation:

So with implicit differentiation, you are going to be taking the derivative of every variable, in the entire equation. Every time you take the derivative of a variable, you have it's rate of change multiplied on the right. In this case, dx or dy. 

The result of the derivative is:

\(\displaystyle 2x*dx+2y*dy=0\)

The first step is to create the \(\displaystyle \frac{dy}{dx}\) term that you are looking to solve for. This is done by dividing the entire equation by \(\displaystyle dx\) to get it on the bottom of the fraction. After distributing this division to each term, the dx in the first term will cancel with itself, and you will be left with one term that is multiplied by \(\displaystyle \frac{dy}{dx}\). At that point, you want to get the term with \(\displaystyle \frac{dy}{dx}\) onto its own side. This can be accomplished by subtracting the \(\displaystyle 2x\) to the right side of the equation. 

The result so far:

\(\displaystyle 2y*\frac{dy}{dx}=-2x\)

Then to finish getting \(\displaystyle \frac{dy}{dx}\) on its own, you divide \(\displaystyle 2y\) to the right side, ending up with:

 

\(\displaystyle \frac{dy}{dx}=\frac{-x}{y}\)

 

So now looking at the question, we know that \(\displaystyle x=4\), so in order to figure out \(\displaystyle y\) we need to plug \(\displaystyle x\) into the equation.

This gives us \(\displaystyle y=3,-3\).

 

So this gives us two possible answers:

\(\displaystyle \frac{dy}{dx}=\frac{-4}{3}or\frac{4}{3}\)

Example Question #258 : Derivatives

Find the derivative of \(\displaystyle f(x)=ln(x^2+x)\)

Possible Answers:

\(\displaystyle \frac{1}{x^2+x}\)

\(\displaystyle \frac{1}{2x+1}\)

\(\displaystyle \frac{1}{x^2}\)

\(\displaystyle ln(2x+1)\)

\(\displaystyle \frac{2x+1}{x^2+x}\)

Correct answer:

\(\displaystyle \frac{2x+1}{x^2+x}\)

Explanation:

So the derivative of a natural log \(\displaystyle ln(x)\) is always equal to \(\displaystyle \frac{1}{x}\) or one over whatever is inside the natural log. In this case \(\displaystyle x^2+x\) is inside the natural log, so the derivative of \(\displaystyle ln(x^2+x)\) should be:

 

\(\displaystyle \frac{1}{x^2+x}\)

But since the inside of the natural log is a function as well, this is the chain rule and the derivative of the natural log will be multiplied by the derivative of the inside, in this case \(\displaystyle x^2+x\), which is \(\displaystyle 2x+1\)

So the final derivative is \(\displaystyle \frac{2x+1}{x^2+x}\)

Example Question #259 : Derivatives

Compute the derivative of the following expression:

\(\displaystyle y = (\sec5x)^5(tan5x)^3\)

Possible Answers:

\(\displaystyle y' = 5(\sec^55x \tan^25x)(5\tan^25x + 3\sec^25x)\)

\(\displaystyle y' = 5(\sec^35x)(5 + 3\sec^65x\tan^25x)\)

\(\displaystyle y' = 5(\tan^75x)(5\tan^25x + 3\sec^25x)\)

\(\displaystyle y' = 25(\sec^35x\tan^25x)(\tan^25x + \sec^35x)\)

\(\displaystyle y' = (\sec^25x \tan5x)(\tan^35x + 25\sec^55x)\)

Correct answer:

\(\displaystyle y' = 5(\sec^55x \tan^25x)(5\tan^25x + 3\sec^25x)\)

Explanation:

This problem involves using product rule, and chain rule.

By product rule,

\(\displaystyle y' = ((\sec5x)^5)'(\tan5x)^3 + (\sec5x)^5((\tan5x)^3)'\)

Then, using power rule, and ten chain.

\(\displaystyle y' = 5(\sec5x)^4 \cdot (\sec5x)'(\tan5x)^3 + (\sec5x)^5 \cdot 3 \cdot (\tan5x)' \cdot(\tan5x)^2\)

Another application of chain rule to get at the angle,

\(\displaystyle y' = 5(\sec5x)^4 \cdot (\sec5x\tan5x)\cdot(5x)'\cdot(\tan5x)^3 + (\sec5x)^5 \cdot 3 \cdot(\sec^25x) \cdot(5x)' \cdot(\tan5x)^2\)Taking the derivative of the angle and then simplifying, we get

\(\displaystyle y' = 5(\sec^55x \tan^25x)(5\tan^25x + 3\sec^25x)\)

Example Question #260 : Derivatives

Find the first derivative of the function:

\(\displaystyle f=e^{\sec(x)}+\cos(2x)\)

Possible Answers:

\(\displaystyle \sec(x)\tan(x)e^{\sec(x)}-\sin(2x)\)

\(\displaystyle \sec(x)\tan(x)e^{\sec(x)}+2\sin(2x)\)

\(\displaystyle \sec(x)\tan(x)e^{\sec(x)}-2\cos(2x)\)

\(\displaystyle \sec(x)\tan(x)e^{\sec(x)}-2\sin(2x)\)

Correct answer:

\(\displaystyle \sec(x)\tan(x)e^{\sec(x)}-2\sin(2x)\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'=\sec(x)\tan(x)e^{\sec(x)}-2\sin(2x)\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}\sec(x)= \sec(x)\tan(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a\)

Note that the chain rule is used for the exponential function (the secant is the inner function) and for the cosine function (the linear term is the inner function). 

Example Question #471 : Ap Calculus Ab

Find the derivative of the function:

\(\displaystyle f=\ln(\cos(2x))\)

Possible Answers:

\(\displaystyle -2\cot(2x)\)

\(\displaystyle 2\tan(2x)\)

\(\displaystyle -2\tan(2x)\)

\(\displaystyle 2\sec(2x)\)

Correct answer:

\(\displaystyle -2\tan(2x)\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'=\frac{1}{\cos(2x)}\cdot (-2\sin(2x))=-2\tan(2x)\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \ln(x)=\frac{1}{x}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} ax=a\)

Note that the chain rule was used on the natural logarithm derivative and the derivative of the cosine function.

Example Question #82 : Chain Rule And Implicit Differentiation

Compute the second derivative of the following function:

\(\displaystyle y=sin(e^x)\)

Possible Answers:

\(\displaystyle y''=e^{cosx}\)

\(\displaystyle y''=e^xcos(e^x)\)

\(\displaystyle y''=e^x(cos(e^x)- sin(e^x))\)

\(\displaystyle y''=e^x(cos(x)- sin(x))\)

\(\displaystyle y''=e^x(cos(e^x)- e^xsin(e^x))\)

Correct answer:

\(\displaystyle y''=e^x(cos(e^x)- e^xsin(e^x))\)

Explanation:

The first step in finding our second derivative is finding the first.

For our function \(\displaystyle y=sin(e^x)\) we note that it is a composite function, and therefore requires the use of the chain rule. The chain rule states if \(\displaystyle y=f(g(x))\) then \(\displaystyle y'=g'(x)f'(g(x))\).

In our case this becomes:

\(\displaystyle y'=e^xcos(e^x)\)

To find the second derivative we must take the derivative of this function we've now computed. To do this we will require both the chain rule as stated, and the product rule since our function is of the form \(\displaystyle y=f(x)g(x)\), our derivative will be of the form \(\displaystyle y'=f'(x)g(x)+g'(x)f(x)\).

Therefore, our second derivative is:
\(\displaystyle y''=e^xcos(e^x)-e^x\cdot e^xsin(e^x)\)

or equivalently,

\(\displaystyle y''=e^x(cos(e^x)- e^xsin(e^x))\)

 

Example Question #83 : Chain Rule And Implicit Differentiation

Given the relation \(\displaystyle y^4=x^2+1\), find \(\displaystyle \frac{dy}{dx}\).

Possible Answers:

None of the other answers

\(\displaystyle \frac{2y}{x^2}\)

\(\displaystyle \frac{x}{2y^3}\)

\(\displaystyle \frac{y}{x^2}\)

\(\displaystyle \frac{2y^3}{x^2}\)

Correct answer:

\(\displaystyle \frac{x}{2y^3}\)

Explanation:

We start by taking the derivative of both sides of the equation, and proceeding as follows,

\(\displaystyle \frac{d}{dx}(y^4)=\frac{d}{dx}(x^2+1)\).

\(\displaystyle 4y^3 \frac{dy}{dx} = 2x\)

\(\displaystyle \frac{dy}{dx} = \frac{x}{2y^3}\)

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