Question

Please help me with this - Implicit Differentiation (Second Derivative)

Answer 1

Step-by-step explanation

Differentiating both sides of the equation with respect to x:

`(d)/(dx)\left(-2x^3-y^2+y\right)=(d)/(dx)\left(0\right)`
`-6x^2-2y(d)/(dx)\left(y\right)+(d)/(dx)\left(y\right)=0`

Isolating d(y)/dx:

`(d)/(dx)\left(y\right)=(6x^2)/(-2y+1)`

Differentiate again both sides of the equation with respect to x:

`(d)/(dx)\left((6x^2)/(-2y+1)\right)`

Take the constant out:

`=6(d)/(dx)\left((x^2)/(-2y+1)\right)`

Applying the quotient rule:

`=6\cdot ((d)/(dx)\left(x^2\right)\left(-2y+1\right)-(d)/(dx)\left(-2y+1\right)x^2)/(\left(-2y+1\right)^2)`
`=6\cdot (2x\left(-2y+1\right)-\left(-2(d)/(dx)\left(y\right)\right)x^2)/(\left(-2y+1\right)^2)`
`=(6\left(2x\left(-2y+1\right)+2x^2(d)/(dx)\left(y\right)\right))/(\left(-2y+1\right)^2)`

The expression now is as follows:

`(d^2)/(dx^2)\left(y\right)=(6\left(2x\left(-2y+1\right)+2x^2(d)/(dx)\left(y\right)\right))/(\left(-2y+1\right)^2)`

Substituting d(y)/dx= 6x^2/(-2y+1):

`(d^2)/(dx^2)\left(y\right)=(6\left(2x\left(-2y+1\right)+2x^2(d)/(dx)\left(y\right)\right))/(\left(-2y+1\right)^2)`
`(d^2)/(dx^2)\left(y\right)=(6\left(2x\left(-2y+1\right)+2x^2\left((6x^2)/(-2y+1)\right)\right))/(\left(-2y+1\right)^2)`

Simplifying the expression by removing the parentheses:

`=(6\left(2x\left(-2y+1\right)+2x^2(6x^2)/(-2y+1)\right))/(\left(-2y+1\right)^2)`
`=(6\left(2x\left(-2y+1\right)+(12x^4)/(-2y+1)\right))/(\left(-2y+1\right)^2)`

Convert the element to a fraction:

`=(2x\left(-2y+1\right)\left(-2y+1\right))/(-2y+1)+(12x^4)/(-2y+1)`

Since the denominators are equal, combine the fractions:

`=(2x\left(-2y+1\right)\left(-2y+1\right)+12x^4)/(-2y+1)`
`=(2x\left(-2y+1\right)^2+12x^4)/(-2y+1)`
`=(2x\left(-2y+1\right)^2+12x^4)/(-2y+1)`
`=(6\cdot (12x^4+2x\left(-2y+1\right)^2)/(-2y+1))/(\left(-2y+1\right)^2)`

Simplifying:

`(d^2)/(dx^2)\left(y\right)=(6\left(2x\left(-2y+1\right)^2+12x^4\right))/(\left(-2y+1\right)^3)`

The final implicit derivative is as follows:

`(d^2)/(dx^2)\left(y\right)=(6\left(2x\left(-2y+1\right)^2+12x^4\right))/(\left(-2y+1\right)^3)`

Now, we need to find the implicit derivative at point (-1,2):

`(d^(2))/(dx^(2))(y)=(6(2*(-1)(-2*(2)+1)^2+12(-1)^4))/((-2*2+1)^3)`
`(d^(2))/(dx^(2))(y)=(6(-2(-4+1)^2+12*1))/((-4+1)^3)`

Adding numbers:

`(d^(2))/(dx^(2))(y)=(6(-2(-3)^2+12))/((-3)^3)`

Computing the powers:

`(d^(2))/(dx^(2))(y)=(6(-18+12))/(-27)`

Adding numbers:

`(d^(2))/(dx^(2))(y)=(6(-6))/(-27)`

Multiplying numbers:

`(d^(2))/(dx^(2))(y)=(-36)/(-27)`

Simplifying:

`(d^(2))/(dx^(2))(y)=(4)/(3)`

The value of the Implicit Derivative at the point (-1,2) is 4/3