AP Calculus AB : Chain rule and implicit differentiation

Study concepts, example questions & explanations for AP Calculus AB

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Example Questions

Example Question #1 : Chain Rule And Implicit Differentiation

\(\displaystyle f(x)=6^(^5^x^{^{2}-7x}^)\).  Find the derivative.

Possible Answers:

\(\displaystyle f'(x)=(30x^2-42x)*ln6\)

\(\displaystyle f'(x)=6^(^5^{x^{2}-7x)}*(10x-7)\)

\(\displaystyle f'(x)=ln6*(10x-7)\)

\(\displaystyle f'(x)=6^(^1^0^x^-^7^)*ln6\)

\(\displaystyle f'(x)=6^(^5^{x^{2}-7x)}*(10x-7)*ln6\)

Correct answer:

\(\displaystyle f'(x)=6^(^5^{x^{2}-7x)}*(10x-7)*ln6\)

Explanation:

When the function is a constant to the power of a function of x, the first step in chain rule is to rewrite f(x).  So, the first factor of f(x) will be \(\displaystyle (6^{5x^2-7x})\).  Next, we have to take the derivative of the function that is the exponent, or \(\displaystyle 5x^2-7x\).  Its derivative is 10x-7, so that is the next factor of our derivative.  Last, when a constant is the base of an exponential function, we must always take the natural log of that number in our derivative.  So, our final factor will be \(\displaystyle ln6\).  Thus, the derivative of the entire function will be all these factors multiplied together: \(\displaystyle f'(x)=6^{5x^2-7x}*(10x-7)*ln6}\).

Example Question #1 : Chain Rule And Implicit Differentiation

Find the derivative of the function: \(\displaystyle y=e^{sinx^3}}\)

Possible Answers:

\(\displaystyle y'=e^{cosx^3}\)

\(\displaystyle y'=e^{sinx^3}*cosx^3\)

\(\displaystyle y'=e^{sinx^3}*cosx^3*3x^2\)

\(\displaystyle y'=sinx^3(e^{sinx^3-1})\)

\(\displaystyle y'=e^{cosx^3}*3x^2\)

Correct answer:

\(\displaystyle y'=e^{sinx^3}*cosx^3*3x^2\)

Explanation:

Whenever we have an exponential function with \(\displaystyle e\), the first term of our derivative will be that term repeated, without changing anything.  So, the first factor of the derivative will be \(\displaystyle e^{sinx^3}\).  Next, we use chain rule to take the derivative of the exponent.  Its derivative is \(\displaystyle cosx^3 * 3x^2\).  So, the final answer is \(\displaystyle y'=e^{sinx^3}*cosx^3*3x^2\).

Example Question #2 : Chain Rule And Implicit Differentiation

Find the derivative of the exponential function, \(\displaystyle y=e^{sec3\theta }\).

Possible Answers:

\(\displaystyle y'=e^{sec(3\theta)tan(3\theta)}\)

\(\displaystyle y'=3e^{sec(3\theta)tan(3\theta)}\)

\(\displaystyle y'=3e^{sec(3\theta)}\)

\(\displaystyle y'=3e^{sec3\theta }*sec(3\theta)tan(3\theta )\)

\(\displaystyle y'=sec(3\theta )tan(3\theta)\)

Correct answer:

\(\displaystyle y'=3e^{sec3\theta }*sec(3\theta)tan(3\theta )\)

Explanation:

To take the derivative of any exponential function, we repeat the exponential function in the derivative.  So, the first factor of the derivative will be \(\displaystyle e^{sec3\theta}\).  Next, we have to take the derivative of the exponent using chain rule.  The derivative of the trigonometric function secx is secxtanx, so in terms of this problem its derivative is \(\displaystyle sec(3\theta)tan(3\theta)\).  Since the angle has a scalar of 3, we must also multiply the entire derivative by 3.  So, the answer is \(\displaystyle y'=3e^{sec3\theta}*sec(3\theta)tan(3\theta)\).

Example Question #1 : Chain Rule And Implicit Differentiation

Find the derivative of \(\displaystyle f(x)=10^{2\sqrt{x}}\).

Possible Answers:

\(\displaystyle f'(x)=10^{2\sqrt{x}}*x^{-{\frac{1}{2}}}\)

\(\displaystyle f'(x)=10^x^{{-{\frac{1}{2}}}}*ln10\)

\(\displaystyle f'(x)=10^{2\sqrt{x}}*ln10*x^{-{\frac{1}{2}}}\)

\(\displaystyle f'(x)=10^{2\sqrt{x}}*2x^{-{\frac{1}{2}}}*ln10\)

\(\displaystyle f'(x)=ln10*x^{-{\frac{1}{2}}}\)

Correct answer:

\(\displaystyle f'(x)=10^{2\sqrt{x}}*ln10*x^{-{\frac{1}{2}}}\)

Explanation:

To find the derivative, we can first rewrite the function to make it easier to take the chain rule.  Rewrite \(\displaystyle 10^{2\sqrt{x}}\) as \(\displaystyle 10^{2x^{\frac{1}{2}}}\).  Now, like in any exponential function, the first factor of the derivative is the original exponential function.  So, the first factor of f'(x) will be \(\displaystyle 10^{2\sqrt{x}}\).  Next, by the chain rule for derivatives, we must take the derivative of the exponent, which is why we rewrote the exponent in a way that is easier to take the derivative of.  So, the derivative of the exponent is \(\displaystyle x^{-{\frac{1}{2}}}\), because the 1/2 and the 2 cancel when we bring the power down front, and the exponent of 1/2 minus 1 becomes negative 1/2.  The last factor of the derivative is \(\displaystyle ln10\) because in every derivative of an exponential function where the base is a number, we must multiply by the natural log of that base.  So, once you multiply all these factors together, the final answer is \(\displaystyle f'(x)=10^{2\sqrt{x}}*ln10*x^{-{\frac{1}{2}}}\)

Example Question #1 : Chain Rule And Implicit Differentiation

If \(\displaystyle x^3+y^3=6xy\) , find the derivative through implicit differentiation.

Possible Answers:

\(\displaystyle y'=\frac{2y+2x}{x-y}\)

\(\displaystyle y'=\frac{y-x^2}{y^2-x}\)

\(\displaystyle y'=\frac{2y-x^2}{y^2-2x}\)

\(\displaystyle y'=\frac{3y-6x^2}{6y^2-3x}\)

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

Correct answer:

\(\displaystyle y'=\frac{2y-x^2}{y^2-2x}\)

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get \(\displaystyle 3x^2+3y^2(y')=6y+6x(y').\)  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: \(\displaystyle 3y^2(y')-6x(y')=6y-3x^2\).  Next, factor out the y' from both terms on the left side of the equation so that we can solve for it: \(\displaystyle y'(3y^2-6x)=6y-3x^2.\)  To get y' alone, divide both sides by \(\displaystyle (3y^2-6x)\) to get \(\displaystyle y'=\frac{6y-3x^2}{3y^2-6x}\).  To simplify even further, we can factor a 2 out of the numerator and denominator and cancel them.  So, the final answer is \(\displaystyle y'=\frac{2y-x^2}{y^2-2x}\).

Example Question #3 : Chain Rule And Implicit Differentiation

If \(\displaystyle sin(x+y)=y^2(cosx)\), find \(\displaystyle y'\).

Possible Answers:

\(\displaystyle y'=\frac{cos(x+y)-y^2(sinx)}{2ycosx-cos(x+y)}\)

\(\displaystyle y'=tanxy-\frac{cos(x+y)}{2ycosx}\)

\(\displaystyle y'=\frac{-2ycosx(sinx)}{cos(x+y)}\)

\(\displaystyle y'=\frac{-2ysinx-cosx}{cos(x+y)-y^2cosx}\)

\(\displaystyle y'=\frac{-y^2(sinx)-cos(x+y)}{cos(x+y)-2ycosx}\)

Correct answer:

\(\displaystyle y'=\frac{-y^2(sinx)-cos(x+y)}{cos(x+y)-2ycosx}\)

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get \(\displaystyle cos(x+y)(1+y')=y^2(-sinx)+cosx(2y*y').\)  The next step is to solve for y', so we put all terms containing y' on the left side of the equation and factor out a common y': \(\displaystyle y'(cos(x+y)-2ycosx)=-y^2sinx-cos(x+y)\).  To get y' alone, divide both sides by\(\displaystyle cos(x+y)-2ycosx\) to get \(\displaystyle y'=\frac{-y^2sinx-cos(x+y)}{cos(x+y)-2ycosx}\).  

Example Question #4 : Chain Rule And Implicit Differentiation

Find the derivative of the function of the circle \(\displaystyle x^2+y^2=25.\)

Possible Answers:

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

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

\(\displaystyle 1\)

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

\(\displaystyle 1\)

Correct answer:

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

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get \(\displaystyle 2x+2y(y')=0.\)  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: \(\displaystyle 2y(y')=-2x.\).  To get y' alone, divide both sides by \(\displaystyle 2y\) to get \(\displaystyle y'=\frac{-2x}{2y}\).  To simplify even further, we can factor a 2 out of the numerator and denominator and cancel them.  So, the final answer is \(\displaystyle \frac{-x}{y}\).

Example Question #5 : Chain Rule And Implicit Differentiation

Find the derivative of the function \(\displaystyle 4x^3-3y=8\) using implicit differentiation.

Possible Answers:

\(\displaystyle y'=\frac{1}{4}x^2\)

cannot be solved

\(\displaystyle y'=\frac{8-4x^3}{-3}\)

\(\displaystyle y'=4x^2\)

\(\displaystyle y'=-4x^2\)

Correct answer:

\(\displaystyle y'=4x^2\)

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get \(\displaystyle 12x^2-3y'=0\).  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: \(\displaystyle -3y'=-12x^2\).  To get y' alone, divide both sides by -3 to get \(\displaystyle y'=\frac{-12x^2}{-3}\).  To simplify even further, we can factor a -2 out of the numerator and denominator and cancel them.  So, the final answer is \(\displaystyle y'=4x^2\).

Example Question #6 : Chain Rule And Implicit Differentiation

Find the derivative of the function \(\displaystyle x^2y+2y^3=3x+2y\).

Possible Answers:

\(\displaystyle y'=\frac{2-3xy}{x^2+2y^2-6}\)

\(\displaystyle y'=\frac{5}{x^2+6y^2}\)

\(\displaystyle y'=\frac{x^2+6y^2-2}{3-2xy}\)

\(\displaystyle y'=\frac{3+2xy}{x^2+2-6y^2}\)

\(\displaystyle y'=\frac{3-2xy}{x^2+6y^2-2}\)

Correct answer:

\(\displaystyle y'=\frac{3-2xy}{x^2+6y^2-2}\)

Explanation:

To find the derivative through implicit differentiation, we have to take the derivative of every term with respect to x.  Don't forget that each time you take the derivative of a term containing y, you must multiply its derivative by y'.  So, when we take the derivative of each term, we get \(\displaystyle 2x(y)+(x^2)y'+(6y^2)y'=3+2y'\).  The next step is to solve for y', so we put all terms containing y' on the left side of the equation: \(\displaystyle x^2y'+6y^2(y')-2y'=3-2xy\).  Next, factor out the y' from both terms on the left side of the equation so that we can solve for it: \(\displaystyle y'(x^2+6y^2-2)=3-2xy\).  To get y' alone, divide both sides by \(\displaystyle (x^2+6y^2-2)\) to get a final answer of  \(\displaystyle y'=\frac{3-2xy}{x^2+6y^2-2}\).  

Example Question #7 : Chain Rule And Implicit Differentiation

Find the derivative of the function \(\displaystyle lny=lnx^6\).

Possible Answers:

\(\displaystyle y'=\frac{6x}{y}\)

\(\displaystyle y'=\frac{6y}{x}\)

undefined

\(\displaystyle y'=\frac{x}{6y}\)

\(\displaystyle y'=\frac{y}{6x}\)

Correct answer:

\(\displaystyle y'=\frac{6y}{x}\)

Explanation:

Before we take the derivative of the logarithmic function, we can make it easier for ourselves by simplifying the equation to \(\displaystyle lny=6lnx\).  We can bring the exponent of 6 down in front of the natural log of x due to properties of logarithms.  Next, take the derivative of each term in terms of x.  Don't forget to multiply by y' each time you take the derivative of a term containing y!  When we do this, we should get \(\displaystyle (\frac{1}{y})y'=6(\frac{1}{x})\) because the derivative of lnx is 1/x.  Next, solve for y' by multiplying both sides by y to get the final answer of \(\displaystyle y'=\frac{6y}{x}\).

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