Question

Implicit functions derivative x^3+y^3=3axy

Answer 1

Answer:

$\Large\boxed{\frac{ay-x^2}{y^2-ax}}$

Step-by-step explanation:

GIVEN:

$\Large\boxed{\textnormal{LESSON: IMPLICIT DERIVATIVE}}$

$\Large\boxed{\textnormal{SUBJECT: MATH}}}$

To solve this problem, first you have to use sum and difference rule by the numbers from left to right.

SOLUTIONS:

First, you have to treat by the y as y(x).

Use differentiate from both sides by using with the x.

$\displaystyle \frac{D}{DX}\left(X^3+Y^3\right)=\frac{D}{DX}\left(3axy\right)$

Used sum/difference rule formula.$\Large\boxed{\textnormal{SUM AND DIFFERENCE RULE FORMULA}}}}}$

$\displaystyle (F\pmG)'=F\:'\pm G'$

$\displaystyle \frac{d}{dx}\left(x^3\right)+\frac{d}{dx}\left(y^3\right)$

Solve.

$\displaystyle \frac{D}{DX}\left(x^3\right)=3x^2$

$\displaystyle \frac{d}{dx}\left(y^3\right)=3y^2\frac{d}{dx}\left(y\right)$

$\displaystyle 3x^2+3y^2\frac{d}{dx}\left(y\right)$

Rewrite the problem down.

$\displaystyle \frac{D}{DX}\left(3axy\right)=3a\left(y+x\frac{D}{DX}\left(y\right)\right)$

Make sure to take out of constant.

$\displaystyle (A*F)'=A*F'$

$\Large\boxed{\textnormal{PRODUCT (MULTIPLY) RULES}}$

$\displaystyle \left(F* G\right)'=F\:'*G+F*G'$

F=x

G=y

$\displaystyle 3a\left(\frac{d}{dx}\left(x\right)y+\frac{d}{dx}\left(y\right)x\right)$

$\displaystyle \frac{D}{DX}\left(X\right)=1$

$\displaystyle 3a\left(1* \:y+\frac{d}{dx}\left(y\right)x\right)$

You have to write convenience of d/dx (y)=y'.

$\displaystyle 3x^2+3y^2y^{'\:}=3a\left(y+xy^{'\:}\right)$

Isolate by the y from one side of the equation.

Expand by using with distributive property.

$\Large\boxed{\textnormal{DISTRIBUTIVE PROPERTY}}}$

$\displaystyle A(B+C)=AB+AC$

3a(y+xy')

A=3a

B=Y

C=XY

$\displaystyle 3ay+3axy^{'\:}$

Rewrite the whole problem down.

$\displaystyle 3x^2+3y^2y^{'\:}=3ay+3axy^{'\:}$

Then, you subtract by 3x² from both sides.

$\displaystyle 3x^2+3y^2y^{'\:}-3x^2=3ay+3axy^{'\:}-3x^2$

Solve.

$\displaystyle 3y^2y^{'\:}=3ay+3axy^{'\:}-3x^2$

Subtract by 3axy' from both sides.

$\displaystyle 3y^2y^{'\:}-3axy^{'\:}=3ay+3axy^{'\:}-3x^2-3axy^{'\:}$

Solve.

$\displaystyle 3y^2y^{'\:}-3axy^{'\:}=3ay-3x^2$

Factor it out.

$\displaystyle 3y^2y^{'\:}-3axy^'$

Common term of 3y'.

$\displaystyle 3y^{'\:}\left(y^2-ax\right)$

Rewrite the problem down again.

$\displaystyle 3y^{'\:}\left(y^2-ax\right)=3ay-3x^2$

Divide by 3(y²-ax) from both sides.

$\displaystyle \frac{3y^{'\:}\left(y^2-ax\right)}{3\left(y^2-ax\right)}=\frac{3ay}{3\left(y^2-ax\right)}-\frac{3x^2}{3\left(y^2-ax\right)}$

Solve. (Simplify to find the answer.)

$\displaystyle y^{'\:}=\frac{ay-x^2}{y^2-ax}$

Don't forget to write by the y' of d/dx (y).

$\displaystyle \boxed{\frac{D}{DX}\left(y\right)=\frac{ay-x^2}{y^2-ax}}$

$\Large\boxed{\frac{D}{DX}\left(y\right)=\frac{ay-x^2}{y^2-ax}=\boxed{\frac{ay-x^2}{y^2-ax} }}$

Hence, the correct answer is ay-x²/y²-ax.