Answer :
Answer:
To find the differentials, we must derivate:
a) y = s/(1 + 5s)
remember that when we have a function:
f(x) = h(x)*g(x)
we have that the differentiation rule is:
f'(x) = h'(x)*g(x) + h(x)*g'(x)
using this rule, we have:
[tex]dy/ds = \frac{1}{1 + 5s} - 5*\frac{s}{(1+5s)^2}[/tex]
Then te differential is:
[tex]dy = (\frac{1}{1 + 5s} - 5*\frac{s}{(1+5s)^2})ds[/tex]
b) now the function is:
y = e^(-u)*cos(u)
We use the same rule as before for the derivation:
[tex]dy/du = -1*e^{-u}*cos(u) + (e^{-u}*(-sin(u))[/tex]
Then we have:
[tex]dy = (-e^{-u}*cos(u) - e^{-u}*sin(u))du = -e^{-u}*(cos(u) + sin(u))du[/tex]
The differential of both the provided functions with respect to s and u respectively are expressed below.
- Derivative of first function is 1/(1+5s)².
- Derivative of second function is [tex]-e^{-u}[\cos (u)+\sin(u)][/tex].
What is differentiation?
The differentiation is the mathematical process of obtaining the derivative of a function.
The first function given as,
[tex]y = \dfrac{s}{ ( 1 + 5s )}[/tex]
Use the quotient rule and derivate the above function with respect to s.
[tex]\dfrac{dy}{ds} = \dfrac{\dfrac{d}{ds} [s]\times(1+5s)-s.\dfrac{d}{ds}(1+5s) }{ ( 1 + 5s )^2}\\\dfrac{dy}{ds} = \dfrac{1\times(1+5s)-s.(0+5\times1) }{ ( 1 + 5s )^2}\\[/tex]
Simplify it further,
[tex]\dfrac{dy}{ds} = \dfrac{1+5s-5s}{ ( 1 + 5s )^2}\\\dfrac{dy}{ds} = \dfrac{1}{ ( 1 + 5s )^2}[/tex]
The second function given as,
[tex]y =e^{-u}\cos (u)[/tex]
Use the product rule and derivate the above function with respect to u.
[tex]\dfrac{dy}{du}=\dfrac{d}{du}[e^{-u}]\times\cos (u)+e^{-u}.\dfrac{d}{du}[\cos(u)]\\\dfrac{dy}{du}=[-e^{-u}(1)]\times\cos (u)-e^{-u}.\sin(u)\\\dfrac{dy}{du}=-e^{-u}[\cos (u)+\sin(u)][/tex]
Hence, the differential of both the provided functions with respect to s and u respectively are expressed below.
- Derivative of first function is 1/(1+5s)².
- Derivative of second function is [tex]-e^{-u}[\cos (u)+\sin(u)][/tex].
Learn more about the differentiation here;
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