If # x = (4t^2)/(5t^2+6) #, and # y = t^3 # then find # dy/dx#?
1 Answer
Feb 9, 2018
# dy/dx = (t(5t^2+6)^2) / (16) #
Explanation:
We have:
# x = (4t^2)/(5t^2+6) # , and# y = t^3 #
Differentiating wrt
# dx/dt = ( (5t^2+6)(8t) - (10t)(4t^2) ) / (5t^2+6)^2 #
# \ \ \ \ \ \ = ( 40t^3+48t - 40t^3) / (5t^2+6)^2 #
# \ \ \ \ \ \ = ( 48t) / (5t^2+6)^2 #
# dy/dt = 3t^2 #
Then, By the chain rule, we have:
# dy/dx = (dy//dt)/(dx//dt) #
# \ \ \ \ \ \ = (3t^2) / (( 48t) / (5t^2+6)^2) #
# \ \ \ \ \ \ = ((3t^2)(5t^2+6)^2) / (48t) #
# \ \ \ \ \ \ = (t(5t^2+6)^2) / (16) #
