Question

If y=log(tanx)+log(Cotx) then dy/dx =?

Answer 1

Answer:

dy/dx = 0

Note:

★ log(A×B) = logA + logB

★ log(A/B) = logA – logB

★ log(A^m) = m×(logA)

★ log(1) = 0

★ e^(logA) = A

(when base of log is e)

★ tan∅ = sin∅/cos∅

★ cot∅ = cos∅/sin∅

★ cot∅ = 1/tan∅

★ tan∅•cot∅ = 1

Solution:

We have ;

=> y = log(tanx) + log(cotx)

=> y = log(tanx•cotx)

=> y = log1

=> y = 0

Now,

Differentiating both sides with respect to x , we get ;

=> dy/dx = d(0)/dx

=> dy/dx = 0

Hence,

The required value of dy/dx is 0.

Answer 2

Given:

Given equation is

y=log ( tanx )+log( cotx )

To Find:

Find the value of dy/dx

Solution:

y=log(tanx)+log(Cotx)

dy/dx = d/dx{log(tanx)+log(Cotx)}

by log a + log b = log ab

we get log(tanx) + log (cotx) = log (tanx.cotx)

applying the above property in the above equation,

dy/dx = d/dx{ log (tanx . cotx)}

dy/dx = d/dx( log( 1)}   (tanx . cotx = 1)

dy/dx =d/dx(0)    ( log 1 = 0)

dy/dx = 0

Hence the value of dy/dx is 0.