Logarithmic differentiation is a method of finding derivatives of some complicated functions, using the properties of logarithms.
There are cases in which differentiating the logarithm of a given function is easier than differentiating the function as it is.
Steps to be followed to find derivative using logarithm.
Step 1 :
Take logarithm on both sides of the given equation.
Step 2 :
Use the properties of logarithm.
Step 3 :
Find derivative with respect to x and solve for dy/dx.
Find derivative of each of the following with respect to x.
Example 1 :
y = x^{cosx}
Solution :
y = x^{cosx}
Take logarithm on both sides.
logy = logx^{cosx}
logy = cosxlogx
Find derivative with respect to x.
(1/y)(dy/dx) = cosx(1/x) + logx(-sinx)
dy/dx = y[(cosx/x) - sinxlogx]
dy/dx = x^{cosx}[(cosx/x) - sinxlogx]
Example 2 :
y = x^{logx }+ (log x)^{x}
Solution :
y = x^{logx }+ (log x)^{x}
Let a = x^{logx }and b = (log x)^{x}
a = x^{logx} loga = log(x^{logx}) loga = logxlog x loga = (log x)^{2 }----(1) |
b = (logx)^{x} logb = log[(logx)^{x}] logb = xlog(logx) ----(2) |
In (1), find derivative with respect to x.
(1/a)(da/dx) = 2logx(1/x)
da/dx = 2alogx/x
In (1), find derivative with respect to x.
(1/b)(db/dx) = x(1/logx)(1/x) + log(logx)(1)
db/dx = b[(1/logx) + log(logx)]
y = x^{logx }+ (log x)^{x}
y = a + b
dy/dx = da/dx + db/dx
dy/d = 2alogx/x + b[(1/logx) + log(logx)]
Substitute a = x^{logx }and b = (logx)^{x}.
dy/dx = (2x^{logx}logx)/x] + (logx)^{x}[logx + log(logx)]
Example 3 :
√(xy) = e^{x - y}
Solution :
√(xy) = e^{x - y}
√x√y = e^{x - y}
Take logarithm on both sides.
log(√x√y) = loge^{x - y}
log√x + log√y = (x - y)loge
logx^{1/2} + logy^{1/2} = x - y
(1/2)logx + (1/2)logy = x - y
Find derivative with respect to x.
(1/2)(1/x) + (1/2)(1/y)(dy/dx) = 1 - dy/dx
1/2x + (1/2y)(dy/dx) = 1 - dy/dx
(1/2y)(dy/dx) + dy/dx = 1 - 1/2x
(dy/dx)(1/2y + 1) = (2x - 1)/2x
(dy/dx)(1 + 2y)/2y = (2x - 1)/2x
dy/dx = (y/x)[(2x - 1)/(1 + 2y)]
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