Differentiate each of the following with respect to x.
Example 1 :
y = [sin x cos (x2)]/[x3 + lnx]
Solution :
Take logarithm on both sides
lny = ln[sinxcos(x2)]/[x3 + lnx]
lny = ln[sinxcos(x2)] - ln(x3 + lnx)
lny = ln(sinx) + lncos(x2) - ln(x3 + lnx) ----(1)
Derivative of ln y :
= (1/y)y'
= y'/y
Derivative of ln(sin x) :
= (1/sinx)cosx
= cosx/sinx
= cot x
Derivative of lncos (x2) :
= (1/cosx2) - sinx2(2x)
= -2xtan x2
Derivative of ln(x3 + ln x) :
= [1/(x3 + lnx)][3x2 + 1/x]
= [1/(x3 + lnx)][(3x3 + 1)/x]
= [(3x3 + 1)/x(x3 + lnx)]
(1) :
y'/y = cotx - 2xtanx2 - [(3x3 + 1)/x(x3 + lnx)]
ny' = y[cotx - 2xtanx2 - [(3x3 + 1)/x(x3 + lnx)]]
ny' = ([sinxcos(x2)]/[x3 + lnx])[cotx - 2xtanx2- [(3x3 + 1)/x(x3 + lnx)]]
Example 2 :
y = (x2 + 2)(x + √2)/√(x + 4)
Solution :
lny = ln{[(x2 + 2)(x + √2)]/√(x + 4)
lny = ln(x2 + 2) + ln(x + √2) - ln √(x + 4) ----(1)
Derivative of lny :
= (1/y)y'
= y'/y
Derivative of ln(x2 + 2) :
= (1/(x2 + 2))(2x)
= 2x/(x2+2)
Derivative of ln(x + √2) :
= 1/(x + √2)
Derivative of ln√(x + 4) :
= (1/√(x + 4)) ⋅ 1/2√(x + 4)
= 1/2(x + 4)
(1) :
y'/y = [2x/(x2 + 2)] + 1/(x + √2) + [1/2(x + 4)]
y' = y[2x/(x2 + 2)] + 1/(x + √2) + [1/2(x + 4)]
y' = ((x2 + 2)(x + √2)/√(x + 4))[2x/(x2 + 2)] + 1/(x + √2) +[1/2(x + 4)]
Example 3 :
y = x√x
Solution :
y = x√x
Take logarithm on both sides.
ln y = ln (x√x)
ln y = √x ln (x)
Differentiate with respect to x.
u = √x and u' = 1/2√x
v = lnx and v' = 1/x
(1/y) y' = √x (1/x) + ln (x)(1/2√x)
= (√x/x) + ln x/2√x
= (1/√x) + ln x/2√x
(y'/y) = (2 + ln x)/2√x
y' = x√x(2 + ln x)/2√x
Example 4 :
y = xsinx
Solution :
y = xsinx
Take logarithm on both sides.
ln y = ln xsinx
ln y = sin x ln x
Differentiate with respect to x.
u = sin x and u' = cos x
v = ln x and v' = 1/x
(1/y)y' = sin x (1/x) + ln x(cos x)
y'/y = sin x/x + cos x (ln x)
y' = y[(sin x/x) + cos x(ln x)]
y' = xsinx[(sin x/x) + cos x(ln x)]
Example 5 :
y = (ln x)cosx
Solution :
y = (ln x)cosx
Take logarithm on both sides.
ln y = ln ((ln x)cosx)
ln y = cosx ln (ln x)
Differentiate with respect to x.
(1/y)y' = cos x (1/ln x)(1/x) + ln(ln x)(-sinx)
y' = y[cos x/x ln x) - sin x(ln (ln x)]
y' = (lnx)cosx(cos x/x ln x) - sin x(ln(ln x))
Example 6 :
y = xlog x + (log x)x
Solution :
y = xlog x + (log x)x
Part 1 :
Let y = xlog x
log y = log (xlog x)
log y = log x (log x)
log y = (log x)2
log y = 2(log x)
Differentiating with respect to x, we get
(1/y) dy/dx = 2(1/x)
dy/dx = y(2/x)
Applying the value of y, we get
= xlog x(2/x)
Part 2 :
y = (log x)x
Taking log on both sides
log y = log ((log x)x)
log y = x log (log x)
(1/y)(dy/dx) = x (1/log x)(1/x) + log (log x) (1)
dy/dx = y[1/log x + log (log x)]
Applying the value of y, we get
= (log x)x [1/log x + log (log x)]
Example 7 :
y = tan-1 (6x/(1 - 9x2))
Solution :
y = tan-1 (6x/(1 - 9x2))
y = tan-1 (2(3x)/(1 - (3x)2))
Let 3x = tan θ
tan 2 θ = 2 tan θ / (1 - tan2θ)
y = tan-1 (2 tan θ/(1 - tan2θ))
y = tan-1 (tan 2 θ)
y = 2 θ
deriving value of θ from 3x = tan θ
θ = tan-1(3x)
y = 2 tan-1(3x)
dy/dx = 2 (1/(1 + (3x)2) (3)
dy/dx = 6/(1 + (3x)2
Example 8 :
y = tan-1 (√(1 - cos x)/(1 + cos x))
Solution :
y = tan-1 (√(1 - cos x)/(1 + cos x))
From cos 2 x = 1 - 2 sin2 x
2sin2 x = 1 - cos 2x
Dividing θ by 2, we get
2sin2 (x/2) = 1 - cos x ----(1)
From cos 2 x = 2 cos2 x - 1
2 cos2 x/2 = 1 + cos x ----(2)
(1) / (2)
(1 - cos x)/(1 + cos x) = 2sin2 (x/2) / 2 cos2 x/2
(1 - cos x)/(1 + cos x) = tan2 (x/2)
Taking square roots on both sides
√(1 - cos x)/(1 + cos x) = √tan2 (x/2)
√(1 - cos x)/(1 + cos x) = tan (x/2)
Applying the above value, we get
y = tan-1 (√(1 - cos x)/(1 + cos x))
y = tan-1 (tan (x/2))
y = x/2
dy/dx = 1/2
Example 9 :
y = sin-1 (3x - 4x3)
Solution :
y = sin-1 (3x - 4x3)
Let x = sin θ
y = sin-1 (3 sin θ - 4(sin3θ))
y = sin-1 (sin 3θ)
y = 3θ
Deriving value of θ
θ = sin-1(x)
y = 3sin-1(x)
Differentiating with respect to x, we get
dy/dx = 3(1 / √(1 - x2))
dy/dx = 3/√(1 - x2)
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