Differentiate the given function. f(x) = (2x + 1)(3x - 2) f(x) = (x - 5)(1 - 2x)

Question

Differentiate the given function. f(x) = (2x + 1)(3x - 2) f(x) = (x - 5)(1 - 2x) y = 10(3u + 1)(1 - 5u) y = 400(15 - x^2)(3x - 2) f(x) = 1/3(x^5 - 2x^3 + 1)(x - 1/x) f(x) = -3(5x^3 - 2x + 5)(Squareroot x + 2x) y = x + 1/x - 2 y = 2x - 3/5x + 4 f(t) = t/t^2 - 2 f(x) = 1/x - 2 y = 3/x + 5 y = t^2 + 1/1 - t^2 f(x) = x^2 - 3x + 2/2x^2 + 5x - 1 g(x) = (x^2 + x + 1)(4 - x)/2x - 1 f(x) = (2 + 5x)^2 f(x) = (x + 1/x)^2 g(t) = t^2 + Squareroot t/2t + 5 h(x) = x/x^2 - 1 + 4 - x/x^2 + 1 In Exercises 19 through 23, find an equation for the tangent line to the given curve at the point where x = x_0. y = (5x - 1)(4 + 3x); x_0 = 0 y = (x^2 + 3x - 1)(2 - x); x_0 = 1 y = x/2x + 3; x_0 = -1 y = x + 7/5 - 2x; x_0 = 0 y = (3 Squareroot x + x)(2 - x^2); x_0 = 1 In Exercises 24 through 27, find all points on the of the given function where the tangent line is f(x) = (x - 1)(x^2 - 8x + 7) f(x) = (x + 1)(x^2 - x - 2) f(x) = x^2 + x - 1/x^2 - x + 1 In Exercises 28 through 31, find the rate of for the pre scribed value of x_0. y = (x^2 + 2)(x + Squareroot x): x_0 = 4 y = (x^2 + 3)(5 - 2x^3); x_0 = 1 y = 2x - 1/3x + 5; x_0 = 1 y = x + 3/2 - 4x; x_0 = 0 The normal line to the curve y = f(x) at the with coordinates (x_0, f(x_0)) is the line per the tangent line at P. In Exercises 32 through an equation for the normal line to the given the prescribed point. y = x^2 + 3x - 5; (0, -5) y = 2/x - Squareroot x; (1, 1)

Explanation / Answer

y=x/(2x+3)

differentiating aboce w.r.t x

dy/dx =(1/(2x+3)) -2x/(2x+3)2

hence at x0=-1

(dy/dx)x0=-1 =3

also at xo=-1 ; y=-1

therefore equation of line -- (y+1)=3(x+1)

x-y+2=0

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