Find an equation of the tangent line to the graph of f(x) = x sin x when x = 0.

Question

Find an equation of the tangent line to the graph of f(x) = x sin x when x = 0. y = 0 f'(x) = 0 y = x cos x + sin x y = x None of these Find d^2y/dx^2 for y = x + 3/x - 1. 0 -8/(x - 1)^3 -4/(x - 1)^3 8/(x - 1)^3 None of these Find d^2y/dx^2 for y = x + 2/x - 3. 10/(x - 3)^3 0 -10/(x - 3)^3 2/(x - 3)^3 None of these Calculate d^2y/dx^2 if y = 1 - x/2 - x. Find the derivative of x^2 f(x). x[x f'(x) + 2 f(x)] 2x f'(x) x[xf(x) + 2 f'(x)] x^2 f'(x) None of these Find the derivative of 9x^2 f(x). 9x^2 f'(x) 9x [xf(x) + 2 f'(x)] 18x f'(x) 9x[xf'(x) + 2 f(x)] None of these

Explanation / Answer

9) at X=0, y=0*sin0 =0

also slope of line dy/dx =sinx+xcosx at x=0,

dy/dx =0

hence equation of line y=mx+c where m=slope =0, and c=0, as line pass through (0,0)

therefore y=0 is the line

option a is right

10)dy/dx =(1/(x-1)2)*((x-1)*d/dx(x+3) -(X+3)d/dx(x-1)) =(1/(x-1)2)(x-1-x-3) =(1/(x-1)2)(-4)

differentiating again

d2y/d2x =8/(x-1)3 hence option d

11) as above d2y/d2x =10/(x-3)3 hence option a

12) dy/dx =2xf(x)+x2f'(x) =x(2f(x)+xf'(x)) hence option a

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