find d/dx when y=(1+x)/e^x
Answers
Given :
Function
To Find :
dy/dx or y'
Solution :
We have:
Now Differentiate with respect to x , by Quotient rule.
More Formula's Related to the topic :
•Chain rule
Let y=f(t) ,t = g(u) and u =m(x) ,then
Explanation:
Given :
Function
\bf\:y=\dfrac{1+x}{e^x}y=
e
x
1+x
To Find :
dy/dx or y'
{\red{\boxed{\large{\bold{Quotient\:Rule}}}}}
QuotientRule
\sf\:(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2},v\:\neq0(
v
u
)
′
=
v
2
vu
′
−uv
′
,v
=0
Solution :
We have:
\bf\:y=\dfrac{1+x}{e^x}y=
e
x
1+x
Now Differentiate with respect to x , by Quotient rule.
\sf\dfrac{dy}{dx}=\dfrac{(e^x)\frac{d(1+x)}{dx}-(1+x)\frac{d(e^x)}{dx}}{(e^x)^2}
dx
dy
=
(e
x
)
2
(e
x
)
dx
d(1+x)
−(1+x)
dx
d(e
x
)
\sf\dfrac{dy}{dx}=\dfrac{e^x(0+1)-(1+x)e^x}{(e^x)^2}
dx
dy
=
(e
x
)
2
e
x
(0+1)−(1+x)e
x
\sf\dfrac{dy}{dx}=\dfrac{e^x(1)-(1+x)e^x}{(e^x)^2}
dx
dy
=
(e
x
)
2
e
x
(1)−(1+x)e
x
\sf\dfrac{dy}{dx}=\dfrac{e^x-e^x-xe^x}{(e^x)^2}
dx
dy
=
(e
x
)
2
e
x
−e
x
−xe
x
\sf\dfrac{dy}{dx}=\dfrac{-xe^x}{(e^x)^2}
dx
dy
=
(e
x
)
2
−xe
x
\sf\dfrac{dy}{dx}=\dfrac{-x}{e^x}
dx
dy
=
e
x
−x
More Formula's Related to the topic :
1)\sf\frac{d(x^n)}{dx}=nx^{n-1}1)
dx
d(x
n
)
=nx
n−1
2)\sf\frac{d(constant)}{dx}=02)
dx
d(constant)
=0
3)\sf\frac{d(e^x)}{dx}=e^x3)
dx
d(e
x
)
=e
x
•Chain rule
Let y=f(t) ,t = g(u) and u =m(x) ,then
\sf\dfrac{dy}{dx}=\dfrac{dy}{dt}\times\dfrac{dt}{du}\times\dfrac{du}{dx}
dx
dy
=
dt
dy
×
du
dt
×
dx
du