Question

x*3 - 2x*2 y*2 + 5x + y - 5 = 0
Then dy/dx at x = 1, y = 1 is equal to..?
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Answer 1

Given :

Function $\bf\:x{}^{3}-2x{}^{2}\times\:y{}^{2}+5x+y-5=0$

To Find :

$\dfrac{dy}{dx}$ at x=1 and y=1

Formula's:

$1) \: \frac{d(x {}^{n} )}{dx} = nx {}^{n - 1}$

$2) \frac{d(constant)}{dx} = 0$

Solution :

$\bf\:x{}^{3}-2x{}^{2}\times\:y{}^{2}+5x+y-5=0$

Differentiate with respect to x

$3x {}^{2} - (2x {}^{2} \times 2y \frac{dy}{dx} + y {}^{2} \times 4x) + 5 + \frac{dy}{dx} = 0$

$3x {}^{2} - 4x {}^{2} y \frac{dy}{dx} - 4x y {}^{2} + 5 + \frac{dy}{dx} = 0$

$3x {}^{2} - 4xy {}^{2} + 5 = \frac{dy}{dx} (4x {}^{2} y) - \frac{dy}{dx}$

$3x {}^{2} - 4xy {}^{2} + 5 = \frac{dy}{dx}(4xy {}^{2} - 1)$

$\implies \dfrac{dy}{dx} = \dfrac{3x {}^{2} - 4xy {}^{2} + 5 }{4x {}^{2} y - 1}$

$\implies \dfrac{dy}{dx} at (1,1) = \frac{3 - 4 + 5}{4 - 1}$

$\implies \bf \dfrac{dy}{dx} at (1,1)= \dfrac{4}{3}$

which is the required solution!

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More About Differention:

• Chain rule

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

Answer 2

Step-by-step explanation:

First step: Take the derivative of the given Q

Second step: Arrange in such a way that dy/dx is in LHS and remaining part comes on RHS.

Third step: Substitute the value of x and y.

Here is your answer.

Follow these steps in every question and you will become the master of it.

Hope it helped you.