Question

Find by differentiation, the gradient of the curve y=ln(sinx) when x=π/4.

Answer 1

Let f(x) = ln(sin(x))

The gradient of the curve is the derivative wrt x. Using the chain rule for composite functions, we get

f'(x) = cos(x) / sin(x) = cot(x)

The gradient of the curve at the point x = pi/4 is f'(pi/4) = cot(pi/4) = 1

Answer 2

given....y=ln(sinx)  and x=pi/4

lel.........sinx=u

y=lnu

dy/du=1/u

        =1/sinx

du/dx=cosx

by appliying chain rule

dy/dx=dy/du*du/dx

       =1/sinx*cosx

dy/dx=cosx/sinx

hte gradient={cos(pi/4)}/sin(pi/4)

                =cos45/sin45

________________  =1