derivative of under root cotx by first principle Share with your friends Share 3 Neha Sethi answered this Dear student Let f(x)=cotx.Then f(x+h)=cot(x+h)∴ddx(f(x))=limh→0f(x+h)-f(x)h⇒ddx(f(x))=limh→0cot(x+h)-cotxhd⇒dx(f(x))=limh→0cot(x+h)-cotxcot(x+h)+cotxhcot(x+h)+cotx⇒ddx(f(x))=limh→0cot(x+h)-cotxhcot(x+h)+cotx⇒ddx(f(x))=limh→0-sinx+h-xsin(x+h) sinxhcot(x+h)+cotx ∵cotA-cotB=-sin(A-B)sinAsinB⇒ddx(f(x))=limh→0-sinhsin(x+h) sinxhcot(x+h)+cotx⇒ddx(f(x))=limh→0-sinhhcot(x+h)+cotxsin(x+h) sinx⇒ddx(f(x))=-limh→0sinhhlimh→01cot(x+h)+cotxsin(x+h) sinx⇒ddx(f(x))=-1×12cotxsin2x⇒ddx(f(x))=-cosec2x2cotx Regards 9 View Full Answer
