Calculus
posted by C .
Implicit differentiation of 5x^22xy+7^y

Two assumptions
1. you want dy/dx
2. You have the equation 5x^2  2xy + 7y^2 = 0
10x  2x dy/dx  2y + 14y dy/dx = 0
dy/dx( 14y  2x) = 2y  10x
dy/dx = (2y10x)/(14y  2x) = (y  5x)/(7y  x) 
No, the equation is
5x^22xy+7^y
Last term has 7 to the y power.... 
Ouch, nasty term ...
the derivative of 7^y would be
(ln7)(7^y)dy/dx
I am sure you can make the necessary changes . 
Yes...very nasty! Thanks
Respond to this Question
Similar Questions

Calculus 1  implicit differentiation
I do not understand implicit differentiation. One of the problems are: find y' by implicit differentiation xy+2x+3x^2=4. I would appreciate any help that can be offered. 
Calculus 1  implicit differentiation
I do not understand implicit differentiation. One of the problems are: find y' by implicit differentiation xy+2x+3x^2=4. I would appreciate any help that can be offered. 
Calculus 1 Implicit differentiation
Help please help me understand what i am doing wrong in this problem! x^3+3x^2y+y^3=8 find dy/dx using implict differentiation. The answer is x^2+2xy  but i keep getting x^2y x^2+y^2  x^2+y^2 
Calculus
Find dy/dx by implicit differentiation: x^22xy+y^3 = c Thank you! 
Calculus
Use implicit differentiation to find the slope of the tangent line to the curve 2(x^2) + 2xy + 2(y^3)= 18 
Calculus
I am having trouble differentiating x^2 + 2lny = y (solving for dy/dx). I know that I need to do implicit differentiation, but the answer I get is not correct. The answer I got was (2x2xy)/2, but the answer is (2xy)/(y2). How do … 
Calculus
Find dy/dx by implicit differentiation. x^3  3x^2y + 2xy^2 =12 Please show me the work/steps on how to do it. 
Calculus
Find dy/dx by implicit differentiation for x^2  2xy + y^3 = C 
Calculus
Find dy/dx by implicit differentiation for x^2  2xy + y^3 = C 
Calculus
Find y" by implicit differentiation. x^3+y^3 = 1 (x^3)'+(y^3)' = (1)' 3x^2+3y^2(y') = 0 3y^2(y') = 3x^2 y' = 3x^2/3y^2 y' = x^2/y^2 y" = [(y^2)(x^2)'(x^2)(y^2)']/y^4 y" = [(y^2)(2x)(x^2)(2y)]/y^4 y" = [(2xy^2)(2yx^2)]/y^4 …