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

Question

Christopher Agresta · Accepted Answer

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

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.!

Charles Arocha · Answer

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

Evelyn Agard · Answer

undefined To find the derivative of $ y = (4x + 3)^{-1} + (x - 4)^{-2} $, apply the chain rule and the power rule. The derivative is:

$$ y' = -\frac{1}{(4x + 3)^2} -2(x - 4)^{-3} $$

This is the derivative of the given function.