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

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Steve M Share
Feb 9, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{t {\left(5 {t}^{2} + 6\right)}^{2}}{16}$

#### Explanation:

We have:

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

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

$\frac{\mathrm{dx}}{\mathrm{dt}} = \frac{\left(5 {t}^{2} + 6\right) \left(8 t\right) - \left(10 t\right) \left(4 {t}^{2}\right)}{5 {t}^{2} + 6} ^ 2$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{40 {t}^{3} + 48 t - 40 {t}^{3}}{5 {t}^{2} + 6} ^ 2$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{48 t}{5 {t}^{2} + 6} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 3 {t}^{2}$

Then, By the chain rule, we have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy} / \mathrm{dt}}{\mathrm{dx} / \mathrm{dt}}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{3 {t}^{2}}{\frac{48 t}{5 {t}^{2} + 6} ^ 2}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{\left(3 {t}^{2}\right) {\left(5 {t}^{2} + 6\right)}^{2}}{48 t}$

$\setminus \setminus \setminus \setminus \setminus \setminus = \frac{t {\left(5 {t}^{2} + 6\right)}^{2}}{16}$

• 17 minutes ago
• 19 minutes ago
• 21 minutes ago
• 22 minutes ago
• 39 seconds ago
• 4 minutes ago
• 6 minutes ago
• 11 minutes ago
• 14 minutes ago
• 16 minutes ago
• 17 minutes ago
• 19 minutes ago
• 21 minutes ago
• 22 minutes ago