differentiation

294 DIFFERENTIAL CALCULUS
Problem 16.
Find the derivative of y = sec ax.
y = sec ax = 1 (i.e. a quotient). Let u = 1 and
cos ax
v = cos ax
dy
v du
dx = dx − udv dx
v 2
i.e.
=
=
(cos ax)(0) − (1)(−a sin ax)
(cos ax) 2
( )( )
a sin ax 1 sin ax
cos 2 ax = a cos ax cos ax
dy
= a sec ax tan ax
dx
Problem 17. Differentiate y = te2t
2 cos t
te2t
The function is a quotient, whose numerator
2 cos t
is a product.
Let u = te 2t and v = 2 cos t then
du
dt = (t)(2e2t ) + (e 2t )(1) and dv =−2 sin t
dt
Hence dy v du
dx = dx − udv dx
i.e.
dy
dx =
v 2
= (2 cos t)[2te2t + e 2t ] − (te 2t )(−2 sin t)
(2 cos t) 2
= 4te2t cos t + 2e 2t cos t + 2te 2t sin t
4 cos 2 t
= 2e2t [2t cos t + cos t + t sin t]
4 cos 2 t
e2t
2 cos 2 (2t cos t + cos t + t sin t)
t
Problem 18. Determine the ( gradient of the
curve y =
5x
√ )
√3, 3
2x 2 + 4 at the point .
2
Let y = 5x and v = 2x 2 + 4
dy
v du
dx = dx − udv dx
v 2
= 10x2 + 20 − 20x 2
(2x 2 + 4) 2 =
( √ )
√3, 3
At the point , x = √ 3,
2
= (2x2 + 4)(5) − (5x)(4x)
(2x 2 + 4) 2
20 − 10x2
(2x 2 + 4) 2
hence the gradient = dy
dx = 20 − 10(√ 3) 2
[2( √ 3) 2 + 4] 2
=
20 − 30
100
Now try the following exercise.
= − 1
10
Exercise 119 Further problems on differentiating
quotients
In Problems 1 to 5, differentiate the quotients
with respect to the variable.
[ ]
2 cos 3x
−6
1.
x 3
x 4 (x sin 3x + cos 3x)
2.
3.
4.
5.
2x
x 2 + 1
3 √ θ 3
2 sin 2θ
ln 2t
√ t
2xe 4x
sin x
[ 2e
4x
[ 2(1 − x 2 ]
)
(x 2 + 1) 2
[
3 √ ]
θ(3 sin 2θ − 4θ cos 2θ)
4 sin 2 2θ
⎡
1 − 1 ⎤
ln 2t
⎢ 2 ⎥
⎣ √ ⎦
t 3
sin 2 x {(1 + 4x) sin x − x cos x} ]
6. Find the gradient of the curve y = 2x
x 2 − 5 at
the point (2, −4).
[−18]
7. Evaluate dy
dx
at x = 2.5, correct to 3 significant
figures, given y = 2x2 + 3
ln 2x . [3.82]