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

2 Answers
Ratnaker Mehta
Apr 11, 2017

# dy/dx=-{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=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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