Multi-Variable Chain Rule - AP Calculus BC

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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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Question

Find $\frac{\mathrm{dk} }{\mathrm{d} t}$ , where , $x=e^{t^2}$ , $y=4t+\cos(t)$

Answer

To find the given derivative of the function, we must use the multivariable chain rule, which states that

$\frac{\mathrm{d}k }{\mathrm{d} t}=\frac{\mathrm{d} k}{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}+\frac{\mathrm{d} k}{\mathrm{d} y}\cdot \frac{\mathrm{d} y}{\mathrm{d} t}$

So, we find all of these derivatives:

$\frac{\mathrm{d} k}{\mathrm{d} x}=2x+y$ , $\frac{\mathrm{d} x}{\mathrm{d} t}=2te^t$ , $\frac{\mathrm{d} k}{\mathrm{d} y}=x$ , $\frac{\mathrm{d} y}{\mathrm{d} t}=4-\sin(t)$

Plugging this into the formula above, and rewriting x and y in terms of t, we get

$(2e^{t^2}+4t+\cos(t))(2te^t)+e^{t^2}(4-\sin(t))$

which simplified becomes

$4te^{t^2+t}+8t^2e^t+2te^t\cos(t)+4e^{t^2}-e^{t^2}\sin(t)$

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Question

Find $\frac{\mathrm{d} g}{\mathrm{d} t}$ of the following function:

, where , ,

Answer

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

$\frac{\mathrm{d} g}{\mathrm{d} t}= \frac{\mathrm{d} g}{\mathrm{d} x}\cdot \frac{\mathrm{d} x}{\mathrm{d} t}+\frac{\mathrm{d} g}{\mathrm{d} y}\cdot \frac{\mathrm{d} y}{\mathrm{d} t}+\frac{\mathrm{d} g}{\mathrm{d} z}\cdot \frac{\mathrm{d} z}{\mathrm{d} t}$

Our partial derivatives are:

$\frac{\mathrm{d} g}{\mathrm{d} x}= 1$ , $\frac{\mathrm{d} g}{\mathrm{d} y}= 2y$ , $\frac{\mathrm{d} g}{\mathrm{d} z}= 3$ , $\frac{\mathrm{d} x}{\mathrm{d} t}= 2t$ , $\frac{\mathrm{d} y}{\mathrm{d} t}= 2t$ , $\frac{\mathrm{d} z}{\mathrm{d} t}= e^t$

Plugging all of this in - and rewriting any remaining variables in terms of t - we get

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