AP Calculus BC - Multi-Variable Chain Rule

Find $\frac{dz}{dt}$ if , and .

Explanation

Find $\frac{dz}{dt}$ if , and .

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.

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

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

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

$e^{t^2}(2t+2t^3)$

$2t^3e^{t^2}$

$4t^4e^{t^2}$

Explanation

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

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

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

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

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

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

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

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

Explanation

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.

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

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

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

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

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

Explanation

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

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, and .

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

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

$\frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

$\frac{dr}{dt}=4t^3-6t^2+4sin\left ( 4t^3-6t^2+4 \right )$

$\frac{dr}{dt}=(-25t^4+48t^3-108t^2+168t-7)sin(5t^5-12t^4+36t^3-84t^2+7t)$

$\frac{dr}{dt}=(5t^4+8t^3-18t^2-9t)cos(4t^5-11t^4+6t^3-8t^2+7t-11)$

$\frac{dr}{dt}=(t^4-3t^3-128t^2-49t)sin(t^5-131t^4+66t^3-98t^2+17t-121)$

Explanation

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

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

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

Find $\frac{dz}{dt}$ if , and .

Explanation

Find $\frac{dz}{dt}$ if , and .

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, and .

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

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

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

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

Find $\frac{dz}{dt}$ if , and .

Explanation

Find $\frac{dz}{dt}$ if , and .

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, and .

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

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

$\frac{dr}{dt}=(2745t^6-486t^5)cos(405t^7-81t^6)$

$\frac{dr}{dt}=(4t^3-6t^2+4)sin\left ( 4t^3-6t^2+4 \right )$

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

$\frac{dr}{dt}=(81t^6)sin(405t^7-81t^6)$

Explanation

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

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

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

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

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

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

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

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

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

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

$\frac{dr}{dt}=\frac{-5}{t(ln(t))}-\frac{4t^2-12t+24}{t(ln(t))^2} + \frac{2t^3-3t}{ln(t)^3}$

$\frac{dr}{dt}=-\frac{7t^4-9t^2+4}{t(ln(t))^4}$

Explanation

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,

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

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

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

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

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

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

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

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