How do you differentiate #f(x) = (sqrtx)/(-x^2-2x+1)# using the quotient rule?
1 Answer
Explanation:
differentiate using the
#color(blue)"quotient rule"# Given
#f(x)=(g(x))/(h(x))" then"#
#color(red)(|bar(ul(color(white)(a/a)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(a/a)|)))........ (A)#
#color(blue)"------------------------------------------------------------"# here
#g(x)=sqrtx=x^(1/2)rArrg'(x)=1/2x^(-1/2)# and
#h(x)=-x^2-2x+1rArrh'(x)=-2x-2#
#color(blue)"------------------------------------------------------------"#
substitute these values into (A)
#f'(x)=((-x^2-2x+1).1/2x^(-1/2)-x^(1/2)(-2x-2))/(-x^2-2x+1)^2# simplifying the numerator.
#=(1/2x^(-1/2)(2(-x^2-2x+1)-2x(-2x-2)))/(-x^2-2x+1)^2#
#=(1/2x^(-1/2)(2x^2+2))/(-x^2-2x+1)^2=(x^(-1/2)(x^2+1))/(-x^2-2x+1)^2#
#rArrf'(x)=(x^2+1)/(sqrtx(-x^2-2x+1)^2)#
