x= 4t^2/5t^2+6, y=t^3 find dy/dx=?

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Steve M Share
Feb 9, 2018

Answer:

# dy/dx = (t(5t^2+6)^2) / (16) #

Explanation:

We have:

# x = (4t^2)/(5t^2+6) #, and # y = t^3 #

Differentiating wrt #t# we have (using the quotient rule):

# 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) #

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