Question

Use logarithmic differentiation to find dy/dx

Answer 1

Answer:

dy/dx  =  (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]

Step-by-step explanation:

y = (x^2 - 3)^sinx

ln y = ln  (x^2 - 3)^sinx

ln y = sin x * ln (x^2 - 3)

1/y * dy/dx  =   sin x * {1 / (x^2 - 3)} * 2x + ln(x^2 - 3) * cos x

1/y dy/dx =  2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)

dy/dx  =   [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)] * y

dy/dx  =  (x^2 - 3)^sin x [2x sin x/ (x^2 - 3) + cos x ln(x^2 - 3)]