EXPLANATION
Differentiating both sides of the equation with respect to x:
[tex]\frac{d}{dx}\left(-2x^3-y^2+y\right)=\frac{d}{dx}\left(0\right)[/tex][tex]-6x^2-2y\frac{d}{dx}\left(y\right)+\frac{d}{dx}\left(y\right)=0[/tex]Isolating d(y)/dx:
[tex]\frac{d}{dx}\left(y\right)=\frac{6x^2}{-2y+1}[/tex]Differentiate again both sides of the equation with respect to x:
[tex]\frac{d}{dx}\left(\frac{6x^2}{-2y+1}\right)[/tex]Take the constant out:
[tex]=6\frac{d}{dx}\left(\frac{x^2}{-2y+1}\right)[/tex]Applying the quotient rule:
[tex]=6\cdot \frac{\frac{d}{dx}\left(x^2\right)\left(-2y+1\right)-\frac{d}{dx}\left(-2y+1\right)x^2}{\left(-2y+1\right)^2}[/tex][tex]=6\cdot \frac{2x\left(-2y+1\right)-\left(-2\frac{d}{dx}\left(y\right)\right)x^2}{\left(-2y+1\right)^2}[/tex][tex]=\frac{6\left(2x\left(-2y+1\right)+2x^2\frac{d}{dx}\left(y\right)\right)}{\left(-2y+1\right)^2}[/tex]The expression now is as follows:
[tex]\frac{d^2}{dx^2}\left(y\right)=\frac{6\left(2x\left(-2y+1\right)+2x^2\frac{d}{dx}\left(y\right)\right)}{\left(-2y+1\right)^2}[/tex]Substituting d(y)/dx= 6x^2/(-2y+1):
[tex]\frac{d^2}{dx^2}\left(y\right)=\frac{6\left(2x\left(-2y+1\right)+2x^2\frac{d}{dx}\left(y\right)\right)}{\left(-2y+1\right)^2}[/tex][tex]\frac{d^2}{dx^2}\left(y\right)=\frac{6\left(2x\left(-2y+1\right)+2x^2\left(\frac{6x^2}{-2y+1}\right)\right)}{\left(-2y+1\right)^2}[/tex]Simplifying the expression by removing the parentheses:
[tex]=\frac{6\left(2x\left(-2y+1\right)+2x^2\frac{6x^2}{-2y+1}\right)}{\left(-2y+1\right)^2}[/tex][tex]=\frac{6\left(2x\left(-2y+1\right)+\frac{12x^4}{-2y+1}\right)}{\left(-2y+1\right)^2}[/tex]Convert the element to a fraction:
[tex]=\frac{2x\left(-2y+1\right)\left(-2y+1\right)}{-2y+1}+\frac{12x^4}{-2y+1}[/tex]Since the denominators are equal, combine the fractions:
[tex]=\frac{2x\left(-2y+1\right)\left(-2y+1\right)+12x^4}{-2y+1}[/tex][tex]=\frac{2x\left(-2y+1\right)^2+12x^4}{-2y+1}[/tex][tex]=\frac{2x\left(-2y+1\right)^2+12x^4}{-2y+1}[/tex][tex]=\frac{6\cdot \frac{12x^4+2x\left(-2y+1\right)^2}{-2y+1}}{\left(-2y+1\right)^2}[/tex]Simplifying:
[tex]\frac{d^2}{dx^2}\left(y\right)=\frac{6\left(2x\left(-2y+1\right)^2+12x^4\right)}{\left(-2y+1\right)^3}[/tex]The final implicit derivative is as follows:
[tex]\frac{d^2}{dx^2}\left(y\right)=\frac{6\left(2x\left(-2y+1\right)^2+12x^4\right)}{\left(-2y+1\right)^3}[/tex]Now, we need to find the implicit derivative at point (-1,2):
[tex]\frac{d^{2}}{dx^{2}}(y)=\frac{6(2*(-1)(-2*(2)+1)^2+12(-1)^4)}{(-2*2+1)^3}[/tex][tex]\frac{d^{2}}{dx^{2}}(y)=\frac{6(-2(-4+1)^2+12*1)}{(-4+1)^3}[/tex]Adding numbers:
[tex]\frac{d^{2}}{dx^{2}}(y)=\frac{6(-2(-3)^2+12)}{(-3)^3}[/tex]Computing the powers:
[tex]\frac{d^{2}}{dx^{2}}(y)=\frac{6(-18+12)}{-27}[/tex]Adding numbers:
[tex]\frac{d^{2}}{dx^{2}}(y)=\frac{6(-6)}{-27}[/tex]Multiplying numbers:
[tex]\frac{d^{2}}{dx^{2}}(y)=\frac{-36}{-27}[/tex]Simplifying:
[tex]\frac{d^{2}}{dx^{2}}(y)=\frac{4}{3}[/tex]The value of the Implicit Derivative at the point (-1,2) is 4/3
