Find derivative of d(b√x) /dt w. r. t. (t). plz sopve my question step by step
Answers
If asked to find the derivative of an integral then you should not evaluate the integral but instead use the fundamental theorem of Calculus.
The FTOC tells us that:
d
d
x
∫
x
a
f
(
t
)
d
t
=
f
(
x
)
for any constant
a
(ie the derivative of an integral gives us the original function back).
Initially we can manipulate the integral as follows (although we have chosen
0
as the lower limit we could in fact choose any constant:
∫
x
2
0
t
3
d
t
=
∫
x
0
t
3
d
t
+
∫
x
2
x
t
3
d
t
And so:
∫
x
2
x
t
3
d
t
=
∫
x
2
0
t
3
d
t
−
∫
x
0
t
3
d
t
And therefore differentiating we get:
d
d
x
∫
x
2
x
t
3
d
t
=
d
d
x
∫
x
2
0
t
3
d
t
−
d
d
x
∫
x
0
t
3
d
t
(notice the upper bounds of the first integral are not in the correct format for the FTOC to be applied, directly). We can manipulate the definite integral using a substitution and the chain rule. Let:
Let:
u
=
x
2
⇒
d
u
d
x
=
2
x
The substituting into the first integral we get:
d
d
x
∫
x
2
x
t
3
d
t
=
d
d
x
∫
u
0
t
3
d
t
−
d
d
x
∫
x
0
t
3
d
t
=
d
u
d
x
⋅
d
d
u
∫
u
0
t
3
d
t
−
d
d
x
∫
x
0
t
3
d
t
=
2
x
d
d
u
∫
u
0
t
3
d
t
−
d
d
x
∫
x
0
t
3
d
t
And now the derivative of the integral is in the correct form for the FTOC to be applied, giving:
d
d
x
∫
x
2
x
t
3
d
t
=
2
x
u
3
−
x
3
=
2
x
(
x
2
)
3
−
x
3
=
2
x
x
6
−
x
3
=
2
x
7
−
x
3