How do you find the logarithmic differentiation of y = x^-cos x ?
first take ln of both sides
ln y = ln(x^-cosx)
ln y = -cosx lnx
now differentiate with respect to x
(dy/dx)/y = -cosx (1/x) + lnx (sinx)
dy/dx = y ( -(1/x)cosx + lnx(sinx))
= x^-cosx ( -(1/x)cosx + lnx(sinx))
To find the logarithmic differentiation of the function y = x^(-cos(x)), follow these steps:
Step 1: Take the natural logarithm (ln) of both sides of the equation:
ln(y) = ln(x^(-cos(x)))
Step 2: Apply the logarithm rule that states ln(a^b) = b * ln(a):
ln(y) = -cos(x) * ln(x)
Step 3: Differentiate both sides of the equation implicitly with respect to x:
d/dx[ln(y)] = d/dx[-cos(x) * ln(x)]
Step 4: Use the chain rule on the left side of the equation:
(1/y) * dy/dx = [-cos(x) * (1/x)] + [sin(x) * ln(x)]
Step 5: Multiply both sides of the equation by y (to eliminate the division):
dy/dx = y * ([sin(x) * ln(x)] - [cos(x)/x])
Step 6: Replace y with its original expression:
dy/dx = x^(-cos(x)) * ([sin(x) * ln(x)] - [cos(x)/x])
So, the logarithmic differentiation of y = x^(-cos(x)) is given by dy/dx = x^(-cos(x)) * ([sin(x) * ln(x)] - [cos(x)/x]).