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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