Question

Y = f(2x-1/x 2 +1) and f '(x) = sinx 2 , then find dy/dx

Answer 1

$y=f(\frac{2x-1}{x^2+1}),\ \ \ f'(x) = Sin\ x^2\\\\use\ chain\ rule:\\\\y'=f'(\frac{2x-1}{x^2+1}) \times \frac{d}{dx}(\frac{2x-1}{x^2+1})\\\\=Sin[(\frac{2x-1}{x^2+1})^2] \times \frac{2(1+x-x^2)}{(x^2+1)^2}$

Answer 2

Answer: y=f(\frac{2x-1}{x^2+1}),\ \ \ f'(x) = Sin\ x^2\\\\use\ chain\ rule:\\\\y'=f'(\frac{2x-1}{x^2+1}) \times \frac{d}{dx}(\frac{2x-1}{x^2+1})\\\\=Sin[(\frac{2x-1}{x^2+1})^2] \times \frac{2(1+x-x^2)}{(x^2+1)^2}

Hope it will help you

Step-by-step explanation: