Chapter 8

8.4 Differentiation of Exponential Functions

Derivative of y = bx,   b > 0, b ≠ 1
y = bx,   b > 0, b ≠ 1
∴ x = logb y

Differentiate both sides with respect to x,
1 = 1⁄y logbe ⋅ dy⁄dx
dy⁄dx = y ⋅ 1⁄logb e = bx logeb = bx ln b

Therefore
d⁄dx bx = bx ln b
Since ln e = 1,
d⁄dx ex = ex
and
d⁄dx eu(x) = eu(x) ⋅ d⁄dx u(x)
In general
d⁄dx bu(x) = bu(x) ⋅ ln b ⋅ d⁄dx u(x)
Example 8.
Differentiate the following functions with respect to x.
(a)   e3x
(b)   e1-x2
(c)   esin x
(d)   x2e3x
(e)   e2x sin 3x
(f)   (ex + e-x)2
(g)   3e2x⁄1 - 2x
Solution
(a)   e3x
d⁄dx e3x = e3x d⁄dx 3x
= e3x ⋅ 3
= 3 e3x

(b)   e1-x2
d⁄dx e1-x2 = e1-x2 d⁄dx (1 - x2)
= e1-x2 ⋅ (-2x)
= -2x ⋅ e1-x2

(c)   esin x
d⁄dx esin x = esin x d⁄dx sin x
= esin x cos x
= cos x ⋅ esin x

(d)   x2e3x
d⁄dx x2e3x = x2 d⁄dx e3x + e3x d⁄dx x2
= x2 ⋅ e3x d⁄dx 3x + e3x ⋅ 2x
= x2 ⋅ e3x ⋅ 3 + e3x ⋅ 2x

(e)   e2x sin 3x
d⁄dx e2x sin 3x = e2x d⁄dx sin 3x + sin 3x d⁄dx e2x
= e2x ⋅ cos 3x d⁄dx 3x + sin 3x ⋅ e2x ⋅ d⁄dx 2x
= e2x ⋅ cos 3x ⋅ 3 + sin 3x ⋅ e2x ⋅ 2
(f)   (ex + e-x)2
d⁄dx (ex + e-x)2 = 2(ex + e-x). d⁄ dx (ex + e-x)
= 2 (ex + e-x).( d⁄ dx ex + d⁄ dx e-x)
= 2 (ex + e-x).(ex + e-x d⁄ dx -x )
= 2 (ex + e-x).(ex + e-x (-1) )
= 2 (ex + e-x).(ex - e-x)

(g) 3.e2x⁄ 1-2x
d⁄ dx 3.e2x⁄ 1-2x
= 3 . d⁄ dx e2x⁄ 1-2x
= 3 . (1 - 2x) d⁄ dx e2x - e2x d⁄ dx (1 - 2x)⁄ (1 - 2x)2
= 3 . (1 - 2x) e2x . d⁄ dx 2x - e2x (-2) ⁄ (1 - 2x)2
= 3 . (1 - 2x) e2x . 2 + 2e2x ⁄ (1 - 2x)2
= 3 . ((1 - 2x).2 + 2) e2x ⁄ (1 - 2x)2
= 3 . (2 - 4x + 2) e2x ⁄ (1 - 2x)2
= 3 . (4 - 4x) e2x ⁄ (1 - 2x)2

Example 8.
Find d⁄dx .
(a) y = ex ln x
(b) y = log10 ex2
(c) y = log3(sin x + ex)
(d) xey + ln (xy) = sin x

Solution
(a) y = ex ln x
dy⁄dx = ex d⁄dx ln x + ln x d⁄dx ex
= ex ⋅ 1⁄x + ln x ⋅ ex

(b) y = log10 ex2 = x2 log10 e
log10 e = 0.4342944819 ဖြစ်သည်။ တနည်းအားဖြင့် constant ပါ။
y = x2 log10 e
dy⁄ dx = x2 d⁄ dx log10 e + log10 e d⁄ dx x2
= x2 . 0 + log10 e . 2x
= log10 e . 2x

(c) y = log3(sin x + ex)
dy⁄ dx = 1⁄ (sin x + ex) .log3 e d⁄ dx (sin x + ex)
= 1⁄ (sin x + ex) .log3 e (cos x + ex)

(d) xey + ln (xy) = sin x

(x d⁄ dx ey + ey dx⁄ dx) + d⁄ dx ln (xy) = d⁄ dx sin x
(x.ey. dy⁄ dx + ey.1) + 1⁄ xy. d⁄ dx (xy) = cos x
(x.ey. dy⁄ dx + ey) + 1⁄ xy⋅(x ⋅ dy⁄ dx + y ⋅ dx⁄ dx) = cos x
(x.ey ⋅ dy⁄ dx + ey) + 1⁄ xy⋅(x⋅ dy⁄ dx + y⋅1) = cos x
x.ey⋅ dy⁄ dx + ey + 1⁄ xy⋅x dy⁄ dx + 1⁄ xy.y = cos x
x.ey ⋅ dy⁄ dx + ey + 1⁄ y ⋅ dy⁄ dx + 1⁄ x = cos x
x.ey ⋅ dy⁄ dx + 1⁄ y. dy⁄ dx = cos x - ey - 1⁄ x
(x.ey + 1⁄ y)⋅ dy⁄ dx = cos x - ey - 1⁄ x
dy⁄ dx = cos x - ey - 1⁄ x ⁄ x.ey + 1⁄ y

Example 10.
Differentiate y = xx,     x > 0.
Solution
y = xx
y = eln xx
y = ex ln x
dy⁄dx = d⁄dx ex ln x
= ex ln x d⁄dx (x ln x)
= ex ln x (x ⋅ 1⁄x + ln x ⋅ 1)
= xx (1 + ln x)

Exercise 8.4
1. Differentiate the following functions with respect to x
(a) (5 + 3x)e-2x
Solution
d ⁄dx (5 + 3x)e-2x = (5 + 3x) d ⁄dx e-2x + e-2x d ⁄dx (5 + 3x)
= (5 + 3x) e-2x d ⁄dx (-2x) + e-2x ⋅ 3
= (5 + 3x) e-2x ⋅ (-2) + e-2x ⋅ 3

(b) 3xx3
တွက်ပုံ
3x ကိုအရင်စဉ်းစားပါမည်။
y = 3x ဖြစ်ပါစေ
နှစ်ဖက်စလုံးကို ln မြှောက်သော်
ln y = ln 3x
ln y = x ln 3 (logb Mk = k logb M နှင့်သဘောတရားတူသည်)
နှစ်ဖက်စလုံးကို x အလိုက် differentiate လုပ်သော်
d⁄ dx ln y = d⁄ dx x ⋅ ln 3
d⁄ dx ln y = x d⁄ dx ln 3 + ln 3 ⋅ dx⁄ dx
1⁄ y dy⁄ dx = x.0 + ln 3.1
1⁄ y. dy⁄ dx = 0 + ln 3
dy⁄ dx = y ln 3
dy⁄ dx = 3x ln 3
∴ d⁄ dx 3x = 3x ln 3
Solution
3x . x3
d⁄ dx 3x ⋅ x3 = 3x d⁄ dx x3 + x3 d⁄ dx 3x
= 3x.3x2 + x3⋅ 3x ln 3 (အပေါ်က အဖြေကိုတန်းထည့်ပါသည်)
= 3x⋅3x2 + x2⋅x ⋅ 3x ln 3
= 3x⋅x2 (3 + x ln 3)

(c) 2x log2 x
d ⁄dx 2x log2 x = 2x d ⁄dx log2 x + log2 x d ⁄dx 2x
= 2x ⋅ 1 ⁄x log2 e + log2 x ⋅ 2x ln 2

(d) 10x log10(x + 1)
d ⁄dx 10x log10(x + 1) = 10x d ⁄dx log10 (x + 1) + log10(x + 1) d ⁄dx 10x
= 10x ⋅ 1 ⁄x+1 log10 e + log10(x+1) ⋅ 10x ln 10

(e) x2 + tan 3x⁄ex
d ⁄dx x2 + tan 3x⁄ex = ex d ⁄dx (x2 + tan 3x) - (x2 + tan 3x) d ⁄dx ex ⁄(ex)2
= ex (2x + sce2 3x d ⁄dx 3x) - (x2 + tan 3x) ⋅ ex ⁄e2x
= ex (2x + sec2 3x ⋅ 3) - (x2 + tan 3x) ⋅ ex ⁄e2x

(f) x ln y + exy = 2

d ⁄dx x ln y + d ⁄dx exy = d ⁄dx 2
(x d ⁄dx ln y + ln y d ⁄dx x) + exy d ⁄dx xy = 0
(x ⋅ 1 ⁄y dy ⁄dx + ln y ⋅ 1) + exy(x d ⁄dx y + y d ⁄dx x) = 0
(x ⁄y dy ⁄dx + ln y) + exy (x⋅ dy ⁄dx + y ⋅ 1) = 0
x ⁄y dy ⁄dx + ln y + x⋅exy dy ⁄dx + y ⋅ exy = 0
x ⁄y dy ⁄dx + x⋅exy dy ⁄dx = - ln y - y ⋅ exy
dy ⁄dx (x ⁄y + x⋅exy) = - ln y - y ⋅ exy
dy ⁄dx = - ln y - y ⋅ exy ⁄ x ⁄y + x⋅exy

2. Given that y = e3x sin 2x, prove that d2y⁄dx2 - 6 dy ⁄dx + 13y = 0
Solution
dy ⁄dx = e3x d ⁄dx sin 2x + sin 2x d ⁄dx e3x
= e3x cos 2x d ⁄dx 2x + sin 2x ⋅ e3x dy ⁄dx 3x
= e3x ⋅ cos 2x ⋅ 2 + sin 2x ⋅ e3x ⋅ 3
= 2 ⋅ e3x ⋅ cos 2x + 3 ⋅ e3x sin 2x
= 2 ⋅ e3x ⋅ cos 2x + 3y

d2y ⁄dx2 = 2[e3x ⋅ d ⁄dx cos 2x + cos 2x ⋅ d ⁄dx e3x] + 3 dy ⁄dx
= 2[e3x (-sin 2x) ⋅ d ⁄dx 2x + cos 2x ⋅ e3x ⋅ d ⁄dx 3x] + 3 dy ⁄dx
= 2[e3x (-sin 2x) ⋅ 2 + cos 2x ⋅ e3x ⋅ 3] + 3 dy ⁄dx
= -4 e3x sin 2x + 6e3x cos 2x + 3(2e3x cos 2x + 3y)
= -4y + 6e3x cos 2x + 6e3x cos 2x + 9y
= 12 e3x cos 2x + 5y
d2y⁄dx2 - 6 dy ⁄dx + 13y = 12 e3x cos 2x + 5y - 6 (2 ⋅ e3x ⋅ cos 2x + 3y) + 13y
= 12 e3x cos 2x + 5y - 12 e3x cos 2x - 18y + 13y
= 0

3. Use logarithmic differentiation to find the derivative of y with respect to x.
$$ \small{\text{(a)} \,\, y = (\sqrt{x})^x } $$ Solution
$$ \small{y = (\sqrt{x})^x} $$ $$ \small{ln \, y = ln \, (\sqrt{x})^x} $$ $$ \small{ = x \, ln \, (\sqrt{x}) } $$ $$ \small{ = x \, \frac{1}{2} \, ln \, x} $$ $$ \small{\frac{d}{dx}\, ln \, y = \frac{d}{dx}\, x\, \frac{1}{2} \, ln \, x} $$ $$ \small{\frac{1}{y} \, \frac{dy}{dx}\, = \frac{1}{2} [x \sdot \frac{d}{dx}\,ln \, x\, + ln \, x \, \frac{d}{dx} \, x]} $$ $$ \small{\frac{1}{y} \, \frac{dy}{dx}\, = \frac{1}{2} [x \sdot \frac{1}{x}\, + ln \, x \, \sdot 1 ]} $$ $$ \small{ = \frac{1}{2} [1\, + ln \, x]} $$ $$ \small{\frac{dy}{dx} = \frac{1}{2} \, y [1\, + ln \, x]} $$ $$ \small{ = \frac{1}{2} \, (\sqrt{x})^x \, (1\, + ln \, x)} $$
(b) y = xcos x
ln y = ln xcos x
= cos x ln x
d ⁄dx ln y = d ⁄dx cos x ln x
1 ⁄y dy ⁄dx = cos x d ⁄dx ln x + ln x d ⁄dx cos x
1 ⁄y dy ⁄dx = cos x ⋅ 1 ⁄x + ln x (-sin x)
dy ⁄dx = y ( cos x ⁄x - sin x ⋅ ln x)
= xcos x (cos x ⁄x - sin x ⋅ ln x)

(c) y = (ln x)ln x
ln y = ln (ln x)ln x
= ln x ⋅ ln (ln x)
d ⁄dx ln y = d ⁄dx [ln x ⋅ ln (ln x)]
1 ⁄y d ⁄dx = ln x ⋅ d ⁄dx ln (ln x) + ln (ln x) d ⁄dx ln x
= ln x ⋅ 1 ⁄ln x ⋅ d ⁄dx ln x + ln (ln x) ⋅ 1 ⁄x
dy ⁄dx = y( 1 ⁄x + ln (ln x) ⁄x)
= (ln x)ln x ⋅ 1 + ln (ln x) ⁄x