Multi-Variable Chain Rule - AP Calculus BC

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Question

Compute $\frac{df}{dt}$ for the function $f(x,y,z)=3xe^{z+y}$ where the variables , and are functions of the of the parameter :

$x = t: : :: : : :y=\ln(t): : : : : : : :z=t^2$

Answer

Compute $\frac{df}{dt}$ for the function $f(x,y,z)=3xe^{z+y}$ where the variables , and are functions of the of the parameter :

$x = t: : :: : : :y=\ln(t): : : : : : : :z=t^2$

Because each function is given in terms of the parameter we can conveniently write the function as a function of alone:

Solution 1

For the function described in this problem, the function can be written strictly in terms of the parameter by simply substituting the definitions given for , or .

$f(x,y,z)=f(t, : \ln t,: :t^2 ) )=3te^{t^2+\ln t}=3t^2e^{t^2}$

$f(t)=3t^2e^{t^2}$

So now we have the function written in the form of a single variable function of We can use the usual chain-rule.

Apply the product rule and and the chain-rule:

$\frac{df}{dt}=3t^2\left(\frac{d}{dt}e^{t^2} \right )+e^{t^2}\left(\frac{d}{dt}3t^2 \right )=6te^{t^2}+3t^2e^{t^2}(2t)$

$=6te^{t^2}+6t^3e^{t^2}$

$=6te^{t^2}(1+t^2)$

Solution 2

The derivative of with respect to will be the sum of the derivatives of each variable , , and with respect to

$\frac{df}{dt}=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}$ (1)

In Equation (1) we use the partial derivative notation for a derivative of with respect to the variables , , and since it is a multi-variable function, we have to specify which variable we are differentiating unless we are differentiating with respect to

We return to the standard derivative notation when computing the derivatives with respect to since each function , , , and can be written as a single variable function of Now simply apply (3) to the function term-by-term:

$\frac{\partial f}{\partial x}\frac{dx}{dt}=3e^{z+y}$

$\frac{\partial f}{\partial y}\frac{dy}{dt}=3xe^{z+y}\frac{1}{t}$

$\frac{\partial f}{\partial z}\frac{dz}{dt}=3xe^{z+y}(2t)=6xte^{z+y}$

Now ad the terms to get an expression for $\frac{df}{dt}$ and then rewrite in terms of

$\frac{df}{dt}=3xe^{z+y}\frac{1}{t}+6xte^{z+y}+3e^{z+y}$

$=3e^{z+y}(1+\frac{x}{t}+2xt)$

$=3e^{t^2+\ln t}\left(1+\frac{t}{t}+2t^2\right)$

$=6e^{t^2+\ln t}\left(1+t^2\right)$

$=6te^{t^2}\left(1+t^2\right)$

Therefore:

$\frac{df}{dt}=6te^{t^2}\left(1+t^2\right)$

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Question

Find the two variable function that satisfies the following:

$\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right )=2$

$\frac{\partial f}{\partial x}(0,1)=1$

Answer

$\frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right )=2$

$\frac{\partial f}{\partial x}(0,1)=1$

The first step is to integrate the function with respect to since that is the outer most derivative in this problem.

$\frac{\partial f}{\partial x}= \int \frac{\partial}{\partial y}\left(\frac{\partial f}{\partial x} \right )dy=\int 2dy=2y+C_1$

$f(x,y)=\int \frac{\partial f}{\partial x}dx=\int (2y+C_1)dx=(2y+C_1)x+C_2$

Now apply the constraints to compute the two constants of integration.

$\frac{\partial f}{\partial x}(0,1)=2+C_1 = 1 : : : : : : :\Rightarrow : : : : : : :C_1=-1$

So far we have:

Now apply the second constraint given:

$f(1,0)=2: : : : : :\ \Rightarrow : : : : : -1+C_2 =2: : : : : \Rightarrow : : : :C_2 =3$

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Question

Compute $\frac{dz}{dt}$ for $z=2xe^{xy}$ , , $y=t^{-2}$ .

Answer

All we need to do is use the formula for multivariable chain rule.

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

$\frac{\partial f}{\partial x}=2e^{xy}+2xye^{xy}$

$\frac{dx}{dt}=3t^2$

$\frac{\partial f}{\partial y}=2x^2e^{xy}$

$\frac{dy}{dt}=-2t^{-3}$

When we put this all together, we get.

$\frac{dz}{dt}=(6t^2e^{xy}+2xye^{xy})-(4t^{-3}x^2e^{xy})$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial z}\frac{dz}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$=\left ( 2z\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$

$=\left ( 2t^2\right )\left ( 2t\right )+\left ( -2\right )\left ( 3t^2-2\right )$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$=\left ( 12a^3\right )\left ( 2t-6\right )+\left ( 10b\right )\left ( 1\right )$

$=\left ( 12\left ( t^2-6t+12\right )^3\right )\left ( 2t-6\right )+\left ( 10\left ( t-5\right )\right )\left ( 1\right )$

$=\left12 ( t^6-18t^5+144t^4-648t^3+1728t^2-2592t+1728\right )\left ( 2t-6\right )+10t-50$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$=-(5t^2-12t+1)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 3t^2+7\right )-(t^3+7t)sin(5t^5-12t^4+36t^3-84t^2+7t)\left ( 10t-12\right )$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$\frac{\partial r}{\partial a}=cos(ab^2)(b^2)$

$\frac{da}{dt}=5$

$\frac{\partial r}{\partial b}=cos(ab^2)(2ab)$

$\frac{db}{dt}=-27t^2$

$\frac{dr}{dt}=\left ( cos(405t^7-81t^6)(81t^6) \right )(5)+(cos(405t^7-81t^6)(-90t^4+18t^3))(-27t^2)$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when $r=e^{x+y}$ , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$\frac{\partial r}{\partial x}=e^{x+y}$

$\frac{dx}{dt}=5+2t$

$\frac{\partial r}{\partial y}=e^{x+y}$

$\frac{dy}{dt}=-7$

$\frac{dr}{dt}=e^{x+y}(5+2t)+e^{x+y}(-7)$

$\frac{dr}{dt}=(2t-2)e^{x+y}$

$\frac{dr}{dt}=(2t-2)e^{5t+t^2+7-7t}=(2t-2)e^{t^2-2t+7}$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when $r=e^{xy}$ , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$\frac{\partial r}{\partial x}=ye^{xy}=(t^2-t-1)e^{xy}$

$\frac{dx}{dt}=-6t^2$

$\frac{\partial r}{\partial y}=xe^{xy}=(-2t^3+1)e^{xy}$

$\frac{dy}{dt}=2t-1$

$\frac{dr}{dt}=(t^2-t-1)e^{xy}(-6t^2)+(-2t^3+1)e^{xy}(2t-1)$

$=(-6t^4+12t^3+6t^2)e^{xy}+(-4t^4+2t^3+2t-1)e^{xy}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{xy}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^3+1)(t^2-t-1)}$

$=(-10t^4+14t^3+6t^2+2t-1)e^{(-2t^5+2t^4+2t^3+t^2-t-1)}$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$\frac{\partial r}{\partial a}=\frac{b}{ab}=\frac{1}{a} =\frac{1}{-2t^2-3t}$

$\frac{da}{dt}=-4t-3$

$\frac{\partial r}{\partial b}=\frac{a}{ab}=\frac{1}{b}=\frac{1}{4t+9}$

$\frac{db}{dt}=4$

$\frac{dr}{dt}=\left ( \frac{1}{-2t^2-3t} \right )(-4t-3)+\left ( \frac{1}{4t+9} \right )(4)$

$=\frac{-4t-3}{-2t^2-3t}+\frac{4}{4t+9}$

$=\frac{(-4t-3)(4t+9)+4(-2t^2-3t)}{(-2t^2-3t)(4t+9)}$

$=\frac{(-4t-3)(4t+9)+4(-2t^2-3t)}{(-2t^2-3t)(4t+9)}$

$\frac{dr}{dt}=\frac{24t^2+60t+27}{8t^3+30t^2+27t}$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial a}\frac{da}{dt}+\frac{\partial r}{\partial b}\frac{db}{dt}$

$\frac{\partial r}{\partial a}=\frac{5}{5a+3b}=\frac{5}{5(8t-3)+3(t^2-7t+1)}=\frac{5}{3t^2+19t-12}$

$\frac{da}{dt}=8$

$\frac{\partial r}{\partial b}=\frac{3}{5a+3b}=\frac{3}{5(8t-3)+3(t^2-7t+1)}=\frac{3}{3t^2+19t-12}$

$\frac{db}{dt}=2t-7$

$\frac{dr}{dt}=\left ( \frac{5}{3t^2+19t-12} \right )(8)+\left ( \frac{3}{3t^2+19t-12} \right )(2t-7)$

$=\left ( \frac{40}{3t^2+19t-12} \right )+\left ( \frac{6t-21}{3t^2+19t-12} \right )$

$\frac{dr}{dt}=\frac{6t+19}{3t^2+19t-12}$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when $r=\frac{x^2-7y^2}{4}$ , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$\frac{\partial r}{\partial x}=\frac{2x}{4}=\frac{x}{2}=\frac{sin(t)}{2}$

$\frac{dx}{dt}=cos(t)$

$\frac{\partial r}{\partial y}=\frac{-14y}{4}=\frac{-7y}{2}=\frac{-7cos(t)}{2}$

$\frac{dy}{dt}=-sin(t)$

$\frac{dr}{dt}=\left (\frac{sin(t)}{2} \right )(cos(t))+\left ( \frac{-7cos(t)}{2} \right )(-sin(t))$

$=\left (\frac{sin(t)cos(t)}{2} \right )+\left ( \frac{7sin(t)cos(t)}{2} \right )$

$\frac{dr}{dt}=4sin(t)cos(t)$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when $r=\frac{4x^2+8y^2}{x^3}$ , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$r=\frac{4x^2+8y^2}{x^3}=\frac{4x^2}{x^3}+\frac{8y^2}{x^3}=\frac{4}{x}+\frac{8y^2}{x^3}$

$\frac{\partial r}{\partial x}=\frac{-4}{x^2}+\frac{(-3)8y^2}{x^4}=\frac{-4}{x^2}-\frac{24y^2}{x^4}=\frac{-4}{(ln(t))^2}-\frac{24(t^2-1)^2}{(ln(t))^4}$

$\frac{\partial r}{\partial x}=\frac{-4}{(ln(t))^2}-\frac{24(t^4-2t^2+1)}{(ln(t))^4}=\frac{-4}{(ln(t))^2}-\frac{24t^4-48t^2+24}{(ln(t))^4}$

$\frac{dx}{dt}=\frac{1}{t}$

$\frac{\partial r}{\partial y}=\frac{16y}{x^3}=\frac{16(t^2-1)}{ln(t)^3}=\frac{16t^2-16}{ln(t)^3}$

$\frac{dy}{dt}=2t$

$\frac{dr}{dt}=\left (\frac{-4}{(ln(t))^2}-\frac{24t^4-48t^2+24}{(ln(t))^4} \right )\left ( \frac{1}{t} \right )+\left ( \frac{16t^2-16}{ln(t)^3}\right )(2t)$

$\frac{dr}{dt}=\frac{-4}{t(ln(t))^2}-\frac{24t^4-48t^2+24}{t(ln(t))^4} + \frac{32t^3-32t}{ln(t)^3}$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when $r=e^{x+4y}$ , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$\frac{\partial r}{\partial x}=e^{x+4y}=e^{4t+4(3t^3-7t)}=e^{12t^3-24t}$

$\frac{dx}{dt}=4$

$\frac{\partial r}{\partial y}=4e^{x+4y}=4e^{4t+4(3t^3-7t)}=4e^{12t^3-24t}$

$\frac{dy}{dt}=9t^2-7$

$\frac{dr}{dt}=\left ( e^{12t^3-24t} \right )(4)+4\left (e^{12t^3-24t} \right )(9t^2-7)$

$\frac{dr}{dt}=(36t^2-24)e^{12t^3-24t}$

$\frac{dr}{dt}=4(9t^2-6)\left (e^{12t^3-24t} \right )$

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Question

Use the chain rule to find $\frac{dr}{dt}$ when $r=\sqrt{x^4-12y}$ , , .

Answer

The chain rule states $\frac{dy}{dx}=\frac{\partial y}{\partial q}\frac{dq}{dx}$ .

Since and are both functions of , $\frac{dr}{dt}$ must be found using the chain rule.

In this problem,

$\frac{dr}{dt}=\frac{\partial r}{\partial x}\frac{dx}{dt}+\frac{\partial r}{\partial y}\frac{dy}{dt}$

$r=\sqrt{x^4-12y}=\left (x^4-12y \right )^{1/2}$

$\frac{\partial r}{\partial x}=\left ( 1/2\right )\left (x^4-12y \right )^{-1/2}\left ( 4x^3\right )=\left ( 2x^3\right )\left (x^4-12y \right )^{-1/2}$

$=\left ( 2(5t^2-12)^3\right )\left ((5t^2-12)^4-12(9-3t^2) \right )^{-1/2}$

$=\frac{250t^6-1800t^4+4320t^2-3456}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

$\frac{dx}{dt}=10t$

$\frac{\partial r}{\partial y}=\left ( 1/2\right )\left (x^4-12y \right )^{-1/2}(-12)=(-6)\left (x^4-12y \right )^{-1/2}$

$=-6\left ((5t^2-12)^4-12(9-3t^2) \right )^{-1/2}$

$=\frac{-6}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

$\frac{dy}{dt}=-6t$

$\frac{dr}{dt}=\left ( \frac{250t^6-1800t^4+4320t^2-3456}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}\right )(10t) +\left (\frac{-6}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}} \right )(-6t)$

$\frac{dr}{dt}=\left ( \frac{2500t^7-18000t^5+43200t^3-34560t}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}\right ) +\left (\frac{36t}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}} \right )$

$\frac{dr}{dt}= \frac{2500t^7-18000t^5+43200t^3-34596t}{\sqrt{625t^8-6000t^6+21600t^4-34524t^2+20628}}$

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Question

Evaluate $\frac{\partial a}{\partial x}$ in terms of and/or if , , , and .

Answer

Expand the equation for chain rule.

$\frac{\partial a}{\partial x} = \frac{\partial a}{\partial x}\frac{\partial x}{\partial r}+ \frac{\partial a}{\partial y}\frac{\partial y}{\partial r} + \frac{\partial a}{\partial z}\frac{\partial z}{\partial r}$

Evaluate each partial derivative.

$\frac{\partial a}{\partial x} = 1$ , $\frac{\partial x}{\partial r} =3r^2$

$\frac{\partial a}{\partial y}= -3$ , $\frac{\partial y}{\partial r} = 2$

$\frac{\partial a}{\partial z} =2z$ , $\frac{\partial z}{\partial r} = 6$

Substitute the terms in the equation.

$\frac{\partial a}{\partial x} =(1)(3r^2)+(-3)(2)+(2z)(6) = 3r^2-6+12z$

Substitute .

The answer is:

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Question

Find $\frac{\mathrm{dg} }{\mathrm{d} t}$ where , , $y=\cos(t)$

Answer

To find the derivative of the function with respect to t, we must use the multivariable chain rule, which states that

$\frac{\mathrm{dg} }{\mathrm{d} t}=\frac{\mathrm{dg} }{\mathrm{d} x}\cdot \frac{\mathrm{dx} }{\mathrm{d} t}$ , where

Using this rule for both variables, we find that

$\frac{\mathrm{dg} }{\mathrm{d} x}=4y$ , $\frac{\mathrm{dg} }{\mathrm{d} y}=4x$ , $\frac{\mathrm{dx} }{\mathrm{d} t}=3t^2$ , $\frac{\mathrm{dy} }{\mathrm{d} t}=-\sin(t)$

Taking the products according to the formula above, and remembering to rewrite x and y in terms of t, we get

$\frac{\mathrm{dg} }{\mathrm{d} t}=12t^2(\cos(t))-4t^3\sin(t)$

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Question

Find $\frac{\mathrm{dh} }{\mathrm{d} t}$ , where and , $y=e^{t^2}$

Answer

To find the derivative of the function with respect to t, we must use the multivariable chain rule, which states that $\frac{\mathrm{dg} }{\mathrm{d} t}=\frac{\mathrm{dg} }{\mathrm{d} x} \cdot \frac{\mathrm{dx} }{\mathrm{d} t}$ for x. (We do the same for the rest of the variables, and add the products together.)

Using the above rule for both variables, we get

$\frac{\mathrm{dh} }{\mathrm{d} x}=2xy$ , $\frac{\mathrm{dx} }{\mathrm{d} t}=1$ , $\frac{\mathrm{dg} }{\mathrm{d} y}=x^2$ , $\frac{\mathrm{dy} }{\mathrm{d} t}=2te^{t^2}$

Plugging all of this into the above formula, and remembering to rewrite x and y in terms of t, we get

$\frac{\mathrm{d} g}{\mathrm{d} t}=e^{t^2}(2t+2t^3)$

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Question

Find $\frac{\mathrm{dj} }{\mathrm{d} t}$ , where $j=xy, x=\cos(t), y=\sin(t)$

Answer

To find the derivative of the function, we must use the multivariable chain rule. For x, this states that $\frac{\mathrm{d}j }{\mathrm{d} t}=\frac{\mathrm{dj} }{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}$ . (We do the same for y and add the results for the total derivative.)

So, our derivatives are:

$\frac{\mathrm{d}j }{\mathrm{d} x}=y$ , $\frac{\mathrm{d} x}{\mathrm{d} t}=-\sin(t)$ , $\frac{\mathrm{d}j }{\mathrm{d} y}=x$ , $\frac{\mathrm{d} y}{\mathrm{d} t}=\cos(t)$

Now, using the above formula and remembering to rewrite x and y in terms of t, we get

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

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Question

Determine $\frac{\mathrm{dg} }{\mathrm{d} t}$ , where , , ,

Answer

To find the derivative of the function with respect to t, we must use the multivariable chain rule, which states that, for x, $\frac{\mathrm{d} g}{\mathrm{d} t}=\frac{\mathrm{dg} }{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}$ . (The same rule applies for the other two variables, and we add the results of the three variables to get the total derivative.)

So, the derivatives are

$\frac{\mathrm{d} g}{\mathrm{d} x}=2x$ , $\frac{\mathrm{d} x}{\mathrm{d} t}=1$ , $\frac{\mathrm{d} g}{\mathrm{d} y}=z$ , $\frac{\mathrm{d} y}{\mathrm{d} t}=e^t$ , $\frac{\mathrm{d} g}{\mathrm{d} z}=y$ , $\frac{\mathrm{d} z}{\mathrm{d} t}=4$

Using the above rule, we get

which rewritten in terms of t, and simplified, becomes

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