The equation of a tangent to the curve y=xf(x^2) at the point with x-coordinate of 1, is given by 2y+3x=6 find f'(1)

Answer 1

Slope of tangent to the curve y = xf(x²) when x = 1 is dy/dx evaluated at x = 1.

By the Product Rule and Chain Rule, dy/dx = f(x²) + xf'(x²)(2x)

So, when x = 1, dy/dx = f(1) + 2f'(1)

The tangent line to the curve y = xf(x²) when x = 1 has equation 2y + 3x = 6, which can also be written as y = (-3/2)x + 3.

So, f(1) + 2f'(1) = -3/2.

Also, when x = 1, y = (-3/2)(1) + 3 = 3/2. The point (1, 3/2) is the point of tangency. So, f(1) = 3/2.

Therefore, 3/2 + 2f'(1) = -3/2

2f'(1) = -3

f'(1) = -3/2