How do you find the derivative of y=(4x+3)^-1+(x-4)^-2y=(4x+3)−1+(x−4)−2?

2 Answers
Ratnaker Mehta
Apr 11, 2017

dy/dx=-{4/(4x+3)^2+2/(x-4)^3}.dydx=−{4(4x+3)2+2(x−4)3}.

Explanation:

In order to avoid the use of Chain Rule in the Diffn., we first

simplify the given fun., as,

y=(4x+3)^-1+(x-4)^-2.y=(4x+3)−1+(x−4)−2.

:. y=1/(4x+3)+1/(x-4)^2=1/(4x+3)+1/(x^2-8x+16)

:. dy/dx=d/dx{1/(4x+3)}+d/dx{1/(x^2-8x+16)}.

:.," by the Quotient Rule, "dy/dx=((4x+3)d/dx(1)-1d/dx(4x+3))/((4x+3)^2)+((x^2-8x+16)d/dx(1)-1d/dx(x^2-8x+16))/(x^2-8x+16)^2.

=(0-1(4))/(4x+3)^2+(0-1(2x-8))/(x^2-8x+16)^2,

=-4/(4x+3)^2-(2(x-4))/((x-4)^2)^2

:. dy/dx=-{4/(4x+3)^2+2/(x-4)^3}.

Enjoy Maths.!

Jim G.
Apr 11, 2017

dy/dx=-[4/(4x+3)^2+2/(x-4)^3]

Explanation:

differentiate each term using the color(blue)"chain rule"

"Given " f(x)=g(h(x))" then "

• f'(x)=g'(h(x))xxh'(x)

dy/dx=-(4x+3)^-2xxd/dx(4x+3)

color(white)(XXXX)-2(x-4)^-3xxd/dx(x-4)

color(white)(dy/dx)=-4(4x+3)^-2-2(x-4)^-3

color(white)(dy/dx)=-[4/(4x+3)^2+2/(x-4)^3]

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