How to Differentiate ? f(t)=((t-2)/(2t+1))^9

1 Answer
Mar 28, 2018

#f^'(t)=45((t-2)^8/((2t+1)^10))#

Explanation:

The quotient rule :
#color(red)(d/(dx)(u/v)=(v(du)/(dx)-u(dv)/(dx))/v^2#

#f(t)=((t-2)/(2t+1))^9#

We know that ,

#color(blue)(d/(dx)(x^n)=nx^(n-1)#

#:.f^'(t)=9((t-2)/(2t+1))^(9-1)d/(dt)((t-2)/(2t+1))...to#[ Chain Rule]

Using the quotient rule

#=>f^'(t)=9((t-2)/(2t+1))^8[((2t+1)(1)-(t-2)(2))/((2t+1)^2)]#

#=>f^'(t)=9((t-2)/(2t+1))^8[(cancel(2t)+1-cancel(2t)+4)/((2t+1)^2)]#

#=>f^'(t)=45((t-2)^8/((2t+1)^10))#