what are the first and second derivatives of the equation: 3x^4y^3-4x^3+7y^2-13

Question

Pretty straight forward, but i am having trouble. You help would be much appreciated

Answer 1

Chris,

 

Did you mean find y' and y'' for

 

3x⁴y³ - 4x³ + 7y² - 13 = 0 ?

 

If so, you can find y'(x,y) by differentiating and separating y':

 

Differentiate both sides:

 

(12x³)y³ + (3x⁴)(3y²y') - 12x² + 14yy' = 0

 

12x³y³ + 9x⁴y²y' - 12x² + 14yy' = 0

 

Combine terms with y' and solve for y' in terms of x and y:

 

y'(9x⁴y² + 14y) = (12x² - 12x³y³)

 

y'(x,y) = (12x² - 12x³y³) / (9x⁴y² + 14y)

 

Now that you have y', just differentiate both sides of this equation using the division rule to obtain y''(x,y,y').  You should be able to take it from here.

Answer 2

you differentiate functions; you did not even provide an equation.

 

If you had one equation involving two unknowns and you cannot solve explicitly for y as a function of x, you can use implicit differentiation--but it requires AN EQUATION, in that case.

 

Revise your question!