Introduction to differentiation 317
i.e.
dy
dx
=
8
3
x^3 +
12
x^4
Problem 10. Iff(t)= 4 t+
1
√
t^3
findf′(t)
f(t)= 4 t+
1
√
t^3
= 4 t+
1
t
3
2
= 4 t^1 +t−
3
2
Hence, f′(t)=( 4 )( 1 )t^1 −^1 +
(
−
3
2
)
t−
3
2 −^1
= 4 t^0 −
3
2
t−
5
2
i.e. f′(t)= 4 −
3
2 t
5
2
= 4 −
3
2
√
t^5
Problem 11. Determine
dy
dx
giveny=
3 x^2 − 5 x
2 x
y=
3 x^2 − 5 x
2 x
=
3 x^2
2 x
−
5 x
2 x
=
3
2
x−
5
2
Hence,
dy
dx
=
3
2
or1.5
Problem 12. Find the differential coefficient of
y=
2
5
x^3 −
4
x^3
+ 4
√
x^5 + 7
y=
2
5
x^3 −
4
x^3
+ 4
√
x^5 + 7
i.e. y=
2
5
x^3 − 4 x−^3 + 4 x
5
(^2) + 7
dy
dx
(
2
5
)
( 3 )x^3 −^1 −( 4 )(− 3 )x−^3 −^1
+( 4 )
(
5
2
)
x
5
2 −^1 + 0
6
5
x^2 + 12 x−^4 + 10 x
3
2
i.e.
dy
dx
6
5
x^2 +
12
x^4
10
√
x^3
Problem 13. Differentiatey=
(x+ 2 )^2
x
with
respect tox
y=
(x+ 2 )^2
x
x^2 + 4 x+ 4
x
x^2
x
4 x
x
4
x
i.e. y=x^1 + 4 + 4 x−^1
Hence,
dy
dx
= 1 x^1 −^1 + 0 +( 4 )(− 1 )x−^1 −^1
=x^0 − 4 x−^2 = 1 −
4
x^2
(sincex^0 = 1 )
Problem 14. Find the gradient of the curve
y= 2 x^2 −
3
x
atx= 2
y= 2 x^2 −
3
x
= 2 x^3 − 3 x−^1
Gradient=
dy
dx
=( 2 )( 2 )x^2 −^1 −( 3 )(− 1 )x−^1 −^1
= 4 x+ 3 x−^2
= 4 x+
3
x^2
Whenx= 2 , gradient= 4 x+
3
x^2
= 4 ( 2 )+
3
( 2 )^2
= 8 +
3
4
=8.75
Problem 15. Find the gradient of the curve
y= 3 x^4 − 2 x^2 + 5 x−2 at the points( 0 ,− 2 )
and( 1 , 4 )
The gradient of a curve at a given point is given by the
corresponding value of the derivative.
Thus, sincey= 3 x^4 − 2 x^2 + 5 x−2,
thegradient=
dy
dx
= 12 x^3 − 4 x+ 5.
At the point( 0 ,− 2 ),x=0, thus
thegradient= 12 ( 0 )^3 − 4 ( 0 )+ 5 = 5
At the point( 1 , 4 ),x=1, thus
thegradient= 12 ( 1 )^3 − 4 ( 1 )+ 5 = 13
Now try the following Practice Exercise
PracticeExercise 133 Differentiation of
y=axnby the general rule (answerson
page 354)
In problems 1 to 20, determine the differential
coefficients with respect to the variable.
- y= 7 x^4 2. y= 2 x+ 1
- y=x^2 −x 4. y= 2 x^3 − 5 x+ 6