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Calculus 3 : Multi-Variable Chain Rule

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Example Questions

Example Question #21 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x+y , x=t^2 and y=2t .

Possible Answers:

t^2

2t^2+2t

2t+2

Correct answer:

2t+2

Explanation:

Find $\frac{dz}{dt}$ if z=x+y , x=t^2 and y=2t .

We use the chain rule to find the total derivative of with respect to .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=1(2t)+1(2)=2t+2$

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Example Question #22 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if $z=e^{x+y}$ , x=lnt and y=1 .

Possible Answers:

$\frac{e^{lnt-1}}{7}$

$\frac{e^{lnt^2}}{4}$

2t^2+2t

$\frac{e^{lnt+1}}{t}$

Correct answer:

$\frac{e^{lnt+1}}{t}$

Explanation:

Find $\frac{dz}{dt}$ if $z=e^{x+y}$ , x=lnt and y=1 .

We use the chain rule to find the total derivative of with respect to .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=\left (e^{x+y}(1) \right )(1/t)+\left (e^{x+y}(1) \right )(0)=\frac{e^{x+y}}{t}$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=lnt and y=1 .

$\frac{dz}{dt}=\frac{e^{lnt+1}}{t}$

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Example Question #23 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x^3y , x=t^2+2t and y=5t-3 .

Possible Answers:

35t^6+162t^5+210t^4+16t^3-72t^2

-5t^6+112t^5-424t^4-26t^3+72t^2

t^6-312t^5+224t^4-46t^3+117t^2

t^6+12t^5+24t^4-6t^3+7t^2

Correct answer:

35t^6+162t^5+210t^4+16t^3-72t^2

Explanation:

Find $\frac{dz}{dt}$ if z=x^3y , x=t^2+2t and y=5t-3 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(y3x^2)(2t+2)+(x^3)(5)=(3x^2y)(2t+2)+5x^3$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=t^2+2t and y=5t-3 .

$\frac{dz}{dt}=\left [ 3(t^2+2t)^2(5t-3) \right ](2t+2)+5(t^2+2t)^3$

$\frac{dz}{dt}=\left [ 3(t^4+4t^3+4t^2)(5t-3) \right ](2t+2)+5(t^6+6t^5+12t^4+8t^3)$

$=\left [ 3(5t^5+20t^4+20t^3-3t^4-12t^3-12t^2) \right ](2t+2)+(5t^6+30t^5+60t^4+40t^3)$

$=\left [ 3(5t^5+17t^4+8t^3-12t^2) \right ](2t+2) +(5t^6+30t^5+60t^4+40t^3)$

=(15t^5+51t^4+24t^3-36t^2)(2t+2) +(5t^6+30t^5+60t^4+40t^3)

=(30t^6+102t^5+48t^4-72t^3 +30t^5+102t^4+48t^3-72t^2) +(5t^6+30t^5+60t^4+40t^3)

=35t^6+162t^5+210t^4+16t^3-72t^2

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Example Question #24 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x^2cosy , x=sin(t) and y=t^2 .

Possible Answers:

2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

cos(t)cos(t^2)-2sin(t^2)

2sin(t)cos(t)-2tsin^2(t)

2sin(t^2)cos(t)cos(t^2)+tsin^2(t^2)sin(t^2)

Correct answer:

2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

Explanation:

Find $\frac{dz}{dt}$ if z=x^2cosy , x=sin(t) and y=t^2 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2xcosy)(cost)+(x^2(-siny))(2t)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=sin(t) and y=t^2 .

$\frac{dz}{dt}=(2sin(t)cos(t^2))(cost)+\left [ (sint)^2(-sin(t^2)) \right ](2t)$

=2sin(t)cos(t)cos(t^2)-2tsin^2(t)sin(t^2)

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Example Question #25 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=ln(xy) , x=cos(t) and y=ln(t) .

Possible Answers:

$tan(2t)-\frac{1}{7lnt}$

$-cos(t)+\frac{1}{t}$

$-tan(t)+\frac{1}{tlnt}$

$sin(t)+\frac{4}{tlnt}$

Correct answer:

$-tan(t)+\frac{1}{tlnt}$

Explanation:

Find $\frac{dz}{dt}$ if z=ln(xy) , x=cos(t) and y=ln(t) .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=\frac{y}{xy}(-sin(t))+\frac{x}{xy}(\frac{1}{t})$

$\frac{dz}{dt}=\frac{1}{x}(-sin(t))+\frac{1}{y}(\frac{1}{t})$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=cos(t) and y=ln(t) .

$\frac{dz}{dt}=\frac{1}{cos(t)}(-sin(t))+\frac{1}{lnt}(\frac{1}{t})$

$\frac{dz}{dt}=\frac{-sin(t)}{cos(t)}+\frac{1}{tlnt}=-tan(t)+\frac{1}{tlnt}$

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Example Question #26 : Multi Variable Chain Rule

Find $\frac{dz}{dt}$ if z=x^2y+cosy , x=-2t^2 and y=sin(t) .

Possible Answers:

$16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

$-56t^3sin(t)+\left [ 3t^4-sin(sin(t)) \right ](sin(t))$

$t^3sin(t)-\left [ sin(cos(t)) \right ](cos(t))$

$6t^3sin(t)+\left [ t^4-cos(sin(t)) \right ](sin(t))$

Correct answer:

$16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

Explanation:

Find $\frac{dz}{dt}$ if z=x^2y+cosy , x=-2t^2 and y=sin(t) .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2xy)(-4t)+(x^2-siny)(cos(t))$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=-2t^2 and y=sin(t) .

$\frac{dz}{dt}=(2(-2t^2)sin(t))(-4t)+\left [ (-2t^2)^2-sin(sin(t)) \right ](cos(t))$

$\frac{dz}{dt}=16t^3sin(t)+\left [ 4t^4-sin(sin(t)) \right ](cos(t))$

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Example Question #381 : Partial Derivatives

Find $\frac{dz}{dt}$ if z=sin(xy) , x=-t^5+4t and y=2t-3 .

Possible Answers:

(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

(-2t^5-5t^4+16t-42)cos(-t^6+8t^2+t^5-12t)

(-12t^5+13t^4+6t-1)sin(-2t^4+4t^2+3t^5-12t)

(4t^5+13t^4+3t+2)sin(-t^6+3t^2+9t^5-2t)

Correct answer:

(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

Explanation:

Find $\frac{dz}{dt}$ if z=sin(xy) , x=-t^5+4t and y=2t-3 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(ycos(xy))(-5t^4+4)+(xcos(xy))(2)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=-t^5+4t and y=2t-3 .

$\frac{dz}{dt}=\left [ (2t-3)cos\left [ (-t^5+4t)(2t-3) \right ] \right ](-5t^4+4) +\left [ (-t^5+4t)cos\left [ (-t^5+4t)(2t-3) \right ] \right ](2)$

$\frac{dz}{dt}=\left [ (2t-3)(-5t^4+4)cos\left [ (-2t^6+8t^2+3t^5-12t) \right ] \right ] +\left [ 2(-t^5+4t)cos\left [ (-2t^6+8t^2+3t^5-12t) \right ] \right ]$

$\frac{dz}{dt}=(-10t^5+8t+15t^4-12)cos(-2t^6+8t^2+3t^5-12t) +(-2t^5+8t)cos(-2t^6+8t^2+3t^5-12t)$ $\frac{dz}{dt}=(-10t^5+8t+15t^4-12-2t^5+8t)cos(-2t^6+8t^2+3t^5-12t)$

=(-12t^5+15t^4+16t-12)cos(-2t^6+8t^2+3t^5-12t)

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Example Question #382 : Partial Derivatives

Find $\frac{dz}{dt}$ if z=xln(y) , x=sin(t) and y=cos(t) .

Possible Answers:

$ln[sin(t)]sin(t)-\frac{cos^2(t)}{sin(t)}$

$ln[cos(t)]cos(t)-\frac{sin^2(t)}{cos(t)}$

$ln[cos(t)]sin(t)-\frac{sin^2(t)}{cos(2t)}$

$ln[sin(t)]cos(t)-\frac{cos^2(t)}{sin(t)}$

Correct answer:

$ln[cos(t)]cos(t)-\frac{sin^2(t)}{cos(t)}$

Explanation:

Find $\frac{dz}{dt}$ if z=xln(y) , x=sin(t) and y=cos(t) .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(ln(y))(cos(t))+(x\frac{1}{y})(-sin(t))$

$\frac{dz}{dt}=ln(y)cos(t)-\frac{x}{y}sin(t)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=sin(t) and y=cos(t) .

$\frac{dz}{dt}=ln(cos(t))cos(t)-\frac{sin(t)}{cos(t)}sin(t)$

$\frac{dz}{dt}=ln[cos(t)]cos(t)-\frac{sin^2(t)}{cos(t)}$

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Example Question #381 : Partial Derivatives

Find $\frac{dz}{dt}$ if z=x^2y-y^2x , x=t^2-5t and y=3t+4 .

Possible Answers:

25t^4 -150t^3 +268t^2 +508t +83

5t^4 +240t^3 -178t^2 -48t +80

15t^4 -140t^3 +168t^2 +408t +80

15t^4 -10t^3 +68t^2 +48t +8

Correct answer:

15t^4 -140t^3 +168t^2 +408t +80

Explanation:

Find $\frac{dz}{dt}$ if z=x^2y-y^2x , x=t^2-5t and y=3t+4 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(2xy-y^2)(2t-5)+(x^2-2yx)(3)$

$\frac{dz}{dt}=(2xy-y^2)(2t-5)+3(x^2-2xy)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, x=t^2-5t and y=3t+4 .

$\frac{dz}{dt}=(2(t^2-5t)(3t+4)-(3t+4)^2)(2t-5)+3((t^2-5t)^2-2(t^2-5t)(3t+4))$

$\frac{dz}{dt}=[2(3t^3-15t^2+4t^2-20t)-(9t^2+24t+16)](2t-5) +3[(t^4-10t^3+25t^2)-2(3t^3-15t^2+4t^2-20t)]$

$\frac{dz}{dt}=[6t^3-30t^2+8t^2-40t-9t^2-24t-16](2t-5) +3[t^4-10t^3+25t^2-6t^3+30t^2-8t^2+40t]$

$\frac{dz}{dt}=[6t^3-31t^2-64t-16](2t-5) +3[t^4-16t^3+47t^2+40t]$

$\frac{dz}{dt}=12t^4-62t^3-128t^2-32t -30t^3+155t^2+320t+80 +3t^4-48t^3+141t^2+120t$

$\frac{dz}{dt}=15t^4 -140t^3 +168t^2 +408t +80$

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Example Question #384 : Partial Derivatives

Find $\frac{dz}{dt}$ if $z=e^{5xy}$ , $x=\sqrt{t}$ and y=t^2 .

Possible Answers:

$\frac{27}{4}t^{5/2}e^{5t^{3/2}}$

$\frac{25}{2}t^{3/2}e^{5t^{5/2}}$

$\frac{25}{4}t^{3/4}e^{5t^{5/4}}$

$\frac{35}{9}t^{3}e^{5t^{5}}$

Correct answer:

$\frac{25}{2}t^{3/2}e^{5t^{5/2}}$

Explanation:

Find $\frac{dz}{dt}$ if $z=e^{5xy}$ , $x=\sqrt{t}$ and y=t^2 .

Keep in mind, when taking the derivative with respect to , is treated as a constant, and when taking the derivative with respect to , is treated as a constant.

$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}$

$\frac{dz}{dt}=(e^{5xy}*5y)(1/2t^{-1/2})+(e^{5xy}*5x)(2t)$

To put $\frac{dz}{dt}$ solely in terms of and , we substitute the definitions of and given in the question, $x=\sqrt{t}$ and y=t^2 .

$\frac{dz}{dt}=(e^{5t^{1/2}(t^2)}*5t^2)(1/2t^{-1/2})+(e^{5t^{1/2}(t^2)}*5t^{1/2})(2t)$

$\frac{dz}{dt}=(e^{5t^{5/2}}*5t^2)(1/2t^{-1/2})+(e^{5t^{5/2}}*5t^{1/2})(2t)$

$\frac{dz}{dt}=(e^{5t^{5/2}}*5/2t^{3/2})+(e^{5t^{5/2}}*10t^{3/2})$

$\frac{dz}{dt}=e^{5t^{5/2}}(5/2t^{3/2}+10t^{3/2})$

$\frac{dz}{dt}=\frac{25}{2}t^{3/2}e^{5t^{5/2}}$

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Calculus 3 : Multi-Variable Chain Rule

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #21 : Multi Variable Chain Rule

Example Question #22 : Multi Variable Chain Rule

Example Question #23 : Multi Variable Chain Rule

Example Question #24 : Multi Variable Chain Rule

Example Question #25 : Multi Variable Chain Rule

Example Question #26 : Multi Variable Chain Rule

Example Question #381 : Partial Derivatives

Example Question #382 : Partial Derivatives

Example Question #381 : Partial Derivatives

Example Question #384 : Partial Derivatives

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