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

Question

1442 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-14T10:50:52

#f'(x)=2(x^2-8)/x^5#

Remembering that:

#d/dxsum_(i=1)^nk_i*f_i(x)=d/dx[k_1f_1(x)+k_2f_2(x)+...+k_nf_n(x)]=#

#=k_isum_(i=1)^nd/dxf_i(x)=k_1f_1'(x)+k_2f_2'(x)+...+k_nf_n'(x)#

And

#d/dx(g(x)/(h(x)))=(h'(x)*g(x)-g'(x)*h(x))/(g^2(x))#

given:

#f(x)=-2-1/x^2+4/x^4#

#:.f'(x)=0-((0*x^2-1*2x)/x^4)+4*((0*x^4-4x^3)/x^8)=#

#=-(-2cancel(x)/(x^(cancel(4)^3)))+4*(-4cancel(x^3)/x^(cancel(8)^5))=#

#=2/x^3-16/x^5=2/x^3(1-8/x^2)=2/x^3(x^2-8)/x^2=2(x^2-8)/x^5#