If z=(x+y)e^y and x=5t and y=1−t^2, find the following derivative using the chain rule. Enter your answer as a function of t.
dz/dt= ?
Answers
By the chain rule, if [tex]z=f(x,y)=f(x(t),y(t))[/tex], then
[tex]\dfrac{\mathrm dz}{\mathrm dt} = \dfrac{\partial f}{\partial x}\dfrac{\mathrm dx}{\mathrm dt} + \dfrac{\partial f}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}[/tex]
We have [tex]f(x,y)=(x+y)e^y[/tex], for which
• [tex]\dfrac{\partial f}{\partial x} = e^y[/tex]
• [tex]\dfrac{\partial f}{\partial y} = e^y + (x+y)e^y = (x+y+1)e^y[/tex]
and [tex]x(t)=5t[/tex] and [tex]y(t)=1-t^2[/tex], so that
• [tex]\dfrac{\mathrm dx}{\mathrm dt} = 5[/tex]
• [tex]\dfrac{\mathrm dy}{\mathrm dt} = -2t[/tex]
Putting everything together, we get
[tex]\dfrac{\mathrm dz}{\mathrm dt} = 5e^y - 2t(x+y+1)e^y \\\\ \dfrac{\mathrm dz}{\mathrm dt} = 5e^{1-t^2} - 2t(5t-t^2+2)e^{1-t^2} = \boxed{\left(2t^3-10t^2-4t+5\right)e^{1-t^2}}[/tex]
The solution of the derivative using the chain rule is [tex]\dfrac{dz}{dt}=(2t^3-10t^2-4t+5)e^{1-t^2}[/tex].
What is the Chain rule?The chain rule explains how to calculate a composite function's derivative.
What is a composite function?If you may write [tex]f(g(x))[/tex] , a function is composite. It's a function within a function, or a function of a function, in other words.
By the chain rule, we can write if [tex]z=f(x,y)=f(x(t),y(t))[/tex], then,
[tex]\dfrac{dz}{dt}=\dfrac{\partial f}{\partial x} \dfrac{dx}{dt}+\dfrac{\partial f}{\partial y} \dfrac{dy}{dt}[/tex]
As it is given to us, [tex]f(x,y)=(x+y)e^y[/tex], therefore, we can write,
[tex]\dfrac{\partial f}{\partial x}=e^y[/tex][tex]\dfrac{\partial f}{\partial y}=e^y+(x+y)e^y=(x+y+1)e^y[/tex]And x(t)=5t, while y(t)=1-t², So,
[tex]\dfrac{dx}{dt}=5[/tex][tex]\dfrac{dy}{dt}=-2t[/tex]Now, Substitute all the values together, in the equation,
[tex]\dfrac{dz}{dt}=\dfrac{\partial f}{\partial x} \dfrac{dx}{dt}+\dfrac{\partial f}{\partial y} \dfrac{dy}{dt}[/tex]
Substituting the values we will get,
[tex]\dfrac{dz}{dt} &=(5 \times e^y)+[-2t(x+y+1)e^y]\\\\[/tex]
[tex]=5e^{1-t^2}-2t(5t-t^2+2)e^{1-t^2}\\\\=(2t^3-10t^2-4t+5)e^{1-t^2}[/tex]
Hence, the solution of the derivative using the chain rule is [tex]\dfrac{dz}{dt}=(2t^3-10t^2-4t+5)e^{1-t^2}[/tex].
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Step-by-step explanation:
First of all, it is helpful to know ...
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Step-by-step explanation:
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