Introduction to differentiation 317
i.e.
dy
dx
=
8
3
x
3
+
12
x 4
Problem 10. If f (t ) = 4t +
1
√
t 3
find f (t )
f (t ) = 4t +
1
√
t 3
= 4t +
1
t
3
2
= 4t
1
+ t
− 3
2
Hence, f
(t ) = (4)(1)t
1−1
+
−
3
2
t
− 3
2 −1
= 4t
0
−
3
2
t
− 5
2
i.e.
f
(t) = 4 −
3
2t
5
2
= 4 −
3
2
√
t 5
Problem 11. Determine
dy
dx
given y =
3x 2 − 5x
2x
y =
3x 2 − 5x
2x
=
3x 2
2x
−
5x
2x
=
3
2
x −
5
2
Hence,
dy
dx
=
3
2
or 1.5
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
− 4x
−3
+ 4x
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
+ 12x
−4
+ 10x
3
2
i.e.
dy
dx
=
6
5
x
2
+
12
x 4 + 10
x 3
Problem 13. Differentiate y =
(x + 2) 2
x
with
respect to x
y =
(x + 2) 2
x
=
x 2 + 4x + 4
x
=
x 2
x
+
4x
x
+
4
x
i.e.
y = x
1
+ 4 + 4x
−1
Hence,
dy
dx
= 1x
1−1
+ 0 + (4)(−1)x
−1−1
= x
0
− 4x
−2
= 1 −
4
x 2
(since x
0
= 1)
Problem 14. Find the gradient of the curve
y = 2x
2
−
3
x
at x = 2
y = 2x
2
−
3
x
= 2x
3
− 3x
−1
Gradient =
dy
dx
= (2)(2)x
2−1
− (3)(−1)x
−1−1
= 4x + 3x
−2
= 4x +
3
x 2
When x = 2, gradient = 4x +
3
x 2 = 4(2) +
3
(2) 2
= 8 +
3
4
= 8.75
Problem 15. Find the gradient of the curve
y = 3x 4 − 2x 2 + 5x − 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, since y = 3x 4 − 2x 2 + 5x − 2,
the gradient =
dy
dx
= 12x 3 − 4x + 5.
At the point (0, −2), x = 0, thus
the gradient = 12(0) 3 − 4(0) + 5 = 5
At the point (1, 4), x = 1, thus
the gradient = 12(1) 3 − 4(1) + 5 = 13
Now try the following Practice Exercise
Practice Exercise 133 Differentiation of
y = ax
n by the general rule (answers on
page 354)
In problems 1 to 20, determine the differential
coefficients with respect to the variable.
1. y = 7x 4
2. y = 2x + 1
3. y = x 2 − x
4. y = 2x 3 − 5x + 6
i.e.
dy
dx
=
8
3
x
3
+
12
x 4
Problem 10. If f (t ) = 4t +
1
√
t 3
find f (t )
f (t ) = 4t +
1
√
t 3
= 4t +
1
t
3
2
= 4t
1
+ t
− 3
2
Hence, f
(t ) = (4)(1)t
1−1
+
−
3
2
t
− 3
2 −1
= 4t
0
−
3
2
t
− 5
2
i.e.
f
(t) = 4 −
3
2t
5
2
= 4 −
3
2
√
t 5
Problem 11. Determine
dy
dx
given y =
3x 2 − 5x
2x
y =
3x 2 − 5x
2x
=
3x 2
2x
−
5x
2x
=
3
2
x −
5
2
Hence,
dy
dx
=
3
2
or 1.5
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
− 4x
−3
+ 4x
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
+ 12x
−4
+ 10x
3
2
i.e.
dy
dx
=
6
5
x
2
+
12
x 4 + 10
x 3
Problem 13. Differentiate y =
(x + 2) 2
x
with
respect to x
y =
(x + 2) 2
x
=
x 2 + 4x + 4
x
=
x 2
x
+
4x
x
+
4
x
i.e.
y = x
1
+ 4 + 4x
−1
Hence,
dy
dx
= 1x
1−1
+ 0 + (4)(−1)x
−1−1
= x
0
− 4x
−2
= 1 −
4
x 2
(since x
0
= 1)
Problem 14. Find the gradient of the curve
y = 2x
2
−
3
x
at x = 2
y = 2x
2
−
3
x
= 2x
3
− 3x
−1
Gradient =
dy
dx
= (2)(2)x
2−1
− (3)(−1)x
−1−1
= 4x + 3x
−2
= 4x +
3
x 2
When x = 2, gradient = 4x +
3
x 2 = 4(2) +
3
(2) 2
= 8 +
3
4
= 8.75
Problem 15. Find the gradient of the curve
y = 3x 4 − 2x 2 + 5x − 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, since y = 3x 4 − 2x 2 + 5x − 2,
the gradient =
dy
dx
= 12x 3 − 4x + 5.
At the point (0, −2), x = 0, thus
the gradient = 12(0) 3 − 4(0) + 5 = 5
At the point (1, 4), x = 1, thus
the gradient = 12(1) 3 − 4(1) + 5 = 13
Now try the following Practice Exercise
Practice Exercise 133 Differentiation of
y = ax
n by the general rule (answers on
page 354)
In problems 1 to 20, determine the differential
coefficients with respect to the variable.
1. y = 7x 4
2. y = 2x + 1
3. y = x 2 − x
4. y = 2x 3 − 5x + 6