If $x = \frac{4 t^{2}}{5 t^{2} + 6}$ $x = (4t^2)/(5t^2+6)$ , and $y = t^{3}$ $y = t^3$ then find $\frac{d y}{d x}$ $dy/dx$ ?

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If $x = \frac{4 t^{2}}{5 t^{2} + 6}$ $x = (4t^2)/(5t^2+6)$ , and $y = t^{3}$ $y = t^3$ then find $\frac{d y}{d x}$ $dy/dx$ ?

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Answer 1 · 2018-02-09T16:55:30

$\frac{d y}{d x} = \frac{t {(5 t^{2} + 6)}^{2}}{16}$ $dy/dx = (t(5t^2+6)^2) / (16)$

Explanation:

We have:

$x = \frac{4 t^{2}}{5 t^{2} + 6}$ $x = (4t^2)/(5t^2+6)$ , and $y = t^{3}$ $y = t^3$

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

$\frac{d x}{d t} = \frac{(5 t^{2} + 6) (8 t) - (10 t) (4 t^{2})}{{(5 t^{2} + 6)}^{2}}$ $dx/dt = ( (5t^2+6)(8t) - (10t)(4t^2) ) / (5t^2+6)^2$

$\ \ \ \ \ \ = ( 40t^3+48t - 40t^3) / (5t^2+6)^2$

$\ \ \ \ \ \ = ( 48t) / (5t^2+6)^2$

$dy/dt = 3t^2$

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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If $x = \frac{4 t^{2}}{5 t^{2} + 6}$ $x = (4t^2)/(5t^2+6)$ , and $y = t^{3}$ $y = t^3$ then find $\frac{d y}{d x}$ $dy/dx$ ?

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If $x = \frac{4 t^{2}}{5 t^{2} + 6}$ $x = (4t^2)/(5t^2+6)$ , and $y = t^{3}$ $y = t^3$ then find $\frac{d y}{d x}$ $dy/dx$ ?