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Higher Order Derivatives

if x = 2at/1+...

Question

if x = 2at/1+t³ and y =2at²/(1+t³)² then dy/ dx =?

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Solution

$x = \frac{2 at}{1 + t^{3}} On differentiating both sides w . r . t t, we get \frac{d x}{d t} = \frac{d (\frac{2 at}{1 + t^{3}})}{d t} = \frac{2 a (1 + t^{3}) - 2 at (3 t^{2})}{{(1 + t^{3})}^{2}} = \frac{2 a + 2 {at}^{3} - 6 {at}^{3}}{{(1 + t^{3})}^{2}} = \frac{2 a - 4 {at}^{3}}{{(1 + t^{3})}^{2}} y = \frac{2 {at}^{2}}{{(1 + t^{3})}^{2}} On differentiating both sides w . r . t t, we get \frac{d y}{d t} = \frac{d \frac{2 {at}^{2}}{{(1 + t^{3})}^{2}}}{d t} = \frac{4 at {(1 + t^{3})}^{2} - 2 {at}^{2} (2 (1 + t^{3}) 3 t^{2})}{{(1 + t^{3})}^{4}} = \frac{4 at {(1 + t^{3})}^{2} - 12 {at}^{4} (1 + t^{3})}{{(1 + t^{3})}^{4}} = \frac{4 at (1 + t^{3}) - 12 {at}^{4}}{{(1 + t^{3})}^{3}} = \frac{4 at + 4 {at}^{4} - 12 {at}^{4}}{{(1 + t^{3})}^{3}} = \frac{4 at - 8 {at}^{4}}{{(1 + t^{3})}^{3}} \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{\frac{4 at - 8 {at}^{4}}{{(1 + t^{3})}^{3}}}{\frac{2 a - 4 {at}^{3}}{{(1 + t^{3})}^{2}}} = \frac{4 at - 8 {at}^{4}}{(1 + t^{3}) (2 a - 4 {at}^{3})} = \frac{4 at (1 - 2 t^{3})}{2 a (1 + t^{3}) (1 - 2 t^{3})} = \frac{2 t}{(1 + t^{3})} ANS . . .$

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