Question

Need help on this as soon as possible I’ve been stuck for awhile

Answer 1

`\textsf{(C)} \quad -(1)/(4)`

Explanation:

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

To find dy/dx for x² + 3y² = 8 + 2xy, differentiate each term with respect to x.

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

`\frac{\text{d}}{\text{d}x}x^2+\frac{\text{d}}{\text{d}x}3y^2=\frac{\text{d}}{\text{d}x}8+\frac{\text{d}}{\text{d}x}2xy`

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

`\implies 2x+\frac{\text{d}}{\text{d}x}3y^2=0+\frac{\text{d}}{\text{d}x}2xy`

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 2x+6y\frac{\text{d}y}{\text{d}x}=0+\frac{\text{d}}{\text{d}x}2xy`

`\implies 2x+6y\frac{\text{d}y}{\text{d}x}=\frac{\text{d}}{\text{d}x}2xy`

Use the product rule to differentiate the term in x and y.

`\textsf{Let}\;u(x)=2x \implies u'(x)=2`

`\textsf{Let}\;v(x)=y \implies v'(x)= 1\frac{\text{d}y}{\text{d}x}`

Therefore:

`\implies \frac{\text{d}}{\text{d}x}[u(x) \cdot v(x)]=u(x) \cdot v'(x)+v(x) \cdot u'(x)`

`\implies \frac{\text{d}}{\text{d}x}2xy=2x \cdot 1 \frac{\text{d}y}{\text{d}x}+y\cdot 2`

`\implies \frac{\text{d}}{\text{d}x}2xy=2x \frac{\text{d}y}{\text{d}x}+2y`

So the final differentiated equation is:

`\implies 2x+6y\frac{\text{d}y}{\text{d}x}=2x \frac{\text{d}y}{\text{d}x}+2y`

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

`\implies 2x+6y\frac{\text{d}y}{\text{d}x}=2x \frac{\text{d}y}{\text{d}x}+2y`

`\implies 6y\frac{\text{d}y}{\text{d}x}-2x \frac{\text{d}y}{\text{d}x}=-2x+2y`

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

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

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

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

To find d²y/dx², differentiate again using the quotient rule and implicit differentiation.

`\textsf{Let}\;u(x)=-x+y \implies u'(x)=-1+1\frac{\text{d}y}{\text{d}x}`

`\textsf{Let}\;v(x)=-x+3y \implies v'(x)=-1+3\frac{\text{d}y}{\text{d}x}`

Therefore:

`\implies \frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}}{\text{d}x}\left[(u(x))/(v(x))\right]=(v(x)\cdot u'(x)-u(x) \cdot v'(x))/([v(x)]^2)`

`\implies \frac{\text{d}^2y}{\text{d}x^2}=\frac{\text{d}}{\text{d}x}\left[(-x+y)/(-x+3y)\right]=\frac{(-x+3y) \cdot \left(-1+1\frac{\text{d}y}{\text{d}x}\right)-(-x+y) \cdot \left(-1+3\frac{\text{d}y}{\text{d}x}\right)}{(-x+3y)^2}`

`\implies \frac{\text{d}^2y}{\text{d}x^2}=\frac{(3y-x)\left(\frac{\text{d}y}{\text{d}x}-1\right)-(y-x)\left(3\frac{\text{d}y}{\text{d}x}-1\right)}{(-x+3y)^2}`

`\implies \frac{\text{d}^2y}{\text{d}x^2}=\frac{\left(3y\frac{\text{d}y}{\text{d}x}-3y-x\frac{\text{d}y}{\text{d}x}+x\right)-\left(3y\frac{\text{d}y}{\text{d}x}-y-3x\frac{\text{d}y}{\text{d}x}+x\right)}{(-x+3y)^2}`

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

Substitute in dy/dx:

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

To find d²y/dx² at (2, 2), substitute x = 2 and y = 2 into the second derivative:

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

`\implies \frac{\text{d}^2y}{\text{d}x^2}=(-4+4\left((0)/(4)\right))/((4)^2)`

`\implies \frac{\text{d}^2y}{\text{d}x^2}=(-4)/(16)`

`\implies \frac{\text{d}^2y}{\text{d}x^2}=-(1)/(4)`