Question

If 3x²-5y²-1=0, find the value of at the point (5,-2). dy dx

Please I really need help ​

Answer 1

`(dy)/(dx)` = -
`(3)/(2)`

Explanation:

differentiating implicitly, using the product rule for the term - 5y²

3x² - 5y² - 1 = 0 ( differentiate implicitly with respect to x )

6x - 10y
`(dy)/(dx)` - 0 = 0 ( subtract 6x from both sides )

- 10y
`(dy)/(dx)` = - 6x ( divide both sides by - 10y )

`(dy)/(dx)` =
`(-6x)/(-10y)` =
`(3x)/(5y)`

substitute (5, - 2 )

`(dy)/(dx)` =
`(3(5))/(5(-2))` =
`(15)/(-10)` = -
`(3)/(2)`

Answer 2

`\frac{\text{d}y}{\text{d}x}=-(3)/(2)`

Explanation:

To differentiate an equation that contains a mixture of x and y terms, use implicit differentiation.

To find dy/dx for 3x² - 5y² - 1 = 0, differentiate each term with respect to x.

Begin by placing d/dx in front of each term of the equation:

`\implies \frac{\text{d}}{\text{d}x} 3x^2-\frac{\text{d}}{\text{d}x} 5y^2-\frac{\text{d}}{\text{d}x} 1=\frac{\text{d}}{\text{d}x} 0`

Differentiate the terms in x only (and constant terms):

`\implies 6x-\frac{\text{d}}{\text{d}x} 5y^2-0=0`

`\implies 6x-\frac{\text{d}}{\text{d}x} 5y^2=0`

Use the chain rule to differentiate terms in y only.

In practice, this means differentiate with respect to y, and place dy/dx at the end:

`\implies 6x-10y\frac{\text{d}y}{\text{d}x}=0`

Rearrange the resulting equation to make dy/dx the subject:

`\implies 10y\frac{\text{d}y}{\text{d}x}=6x`

`\implies \frac{\text{d}y}{\text{d}x}=(6x)/(10y)`

Now we have differentiated the given equation with respect to x, substitute the given point (5, -2) into the differentiated equation:

`\implies \frac{\text{d}y}{\text{d}x}=(6(5))/(10(-2))`

`\implies \frac{\text{d}y}{\text{d}x}=(30)/(-20)`

`\implies \frac{\text{d}y}{\text{d}x}=-(3)/(2)`

Therefore, the value of dy/dx at point (5, -2) is -3/2.