Wikipedia
derivative of implicit functions : This creates two derivatives: one for y > 0 and another for y < 0. One might find it substantially easier to implicitly differentiate the implicit function; ..... More from Wikipedia
Derivative of Implicit Functions
Derivative of Implicit Functions - Till now, the functions that we have discussed, are explicitly functions of x. We have defined y in terms of x. Suppose we have an equation f(x,y) = 0, which cannot be put in the form of y=f(x) to differentiate in ..
Derivative of Implicit Functions
Till now, the functions that we have discussed, are explicitly functions of x. We have defined y in terms of x. Suppose we have an equation f(x,y) = 0, which cannot be put in the form of y=f(x) to differentiate in the usual way, we can still differentiate the equation f(x,..
A hairier implicit differentiation problem.
This video explains how to find the derivative of composite functions involving trigonometric functions. Basic derivatives, implicit differentiation, chain rule, product rule and power rule are used in the examples.
Question : so, on my math homework, there's an equation x^2+xy+y^2=3. we're currently doing derivatives of implicit functions, but i just don't understand how the process of figuring out what the derivative is as it's stated in the book. can someone give me a little help and lay out the steps to solving this with the given equation?
thanks!
Answer : We differentiate both sides separately. I put a d/dx notation so I won't get lost, although you can skip this part and go straight to differentiating. So we have: d/dx(x + xy + y ) = d/dx(3) We differentiate normally. However, since we're differentiating with respect to x, we must attach dy/dx when differentiating y variables only. Take note that we must use the product rule in differentiating xy. Solution: d/dx(x + xy + y ) = d/dx(3) 2x + (1)y + x(1)(dy/dx) + 2y(dy/dx) = 0 2x + y + x(dy/dx) + 2y(dy/dx) = 0 Transpose 2x and y. x(dy/dx) + 2y(dy/dx) = -2x - y (Factor dy/dx out) dy/dx(x + 2y) = -2x - y (Divide both sides by x + 2y) dy/dx = -2x - y / (x + 2y) (Answer) Notice we have now isolated dy/dx on one side of the equation; hence, we have successfully differentiated implicitly. Hope this helps!.. More from Yahoo Answers
Answer : We differentiate both sides separately. I put a d/dx notation so I won't get lost, although you can skip this part and go straight to differentiating. So we have: d/dx(x + xy + y ) = d/dx(3) We differentiate normally. However, since we're differentiating with respect to x, we must attach dy/dx when differentiating y variables only. Take note that we must use the product rule in differentiating xy. Solution: d/dx(x + xy + y ) = d/dx(3) 2x + (1)y + x(1)(dy/dx) + 2y(dy/dx) = 0 2x + y + x(dy/dx) + 2y(dy/dx) = 0 Transpose 2x and y. x(dy/dx) + 2y(dy/dx) = -2x - y (Factor dy/dx out) dy/dx(x + 2y) = -2x - y (Divide both sides by x + 2y) dy/dx = -2x - y / (x + 2y) (Answer) Notice we have now isolated dy/dx on one side of the equation; hence, we have successfully differentiated implicitly. Hope this helps!.. More from Yahoo Answers
Question : When Solving an implicit derivative function is it necessary to simplify first? Or can you take the derivative and then simplify. Here's an example of one I've done just to see if I'm doing it right. ( I took derivatives first)
x^2 + x*y-y^3 = x*y^2 Initial problem
2x + 1y - 3y^2 = 1*2y
2x = 2y - 1y + 3y
2x = 4y
y = 2x/4
Am I missing something or is this all there is to it? Thanks for the help in advance Thanks Luis and Sahsjing, both were pretty easy to follow except for how you de..
Answer : You missed all y' chains. x^2 + x*y-y^3 = x*y^2 Initial problem Taking derivative with respect to x, 2x + x'y + xy' - 3y^2y' = x'y^2+x*(y^2)' 2x + y + xy' - 3y^2y' = y^2 + 2xyy' Solving for y', y' = (2x+y-y^2)/(2xy-x+3y^2).. More from Yahoo Answers
Answer : You missed all y' chains. x^2 + x*y-y^3 = x*y^2 Initial problem Taking derivative with respect to x, 2x + x'y + xy' - 3y^2y' = x'y^2+x*(y^2)' 2x + y + xy' - 3y^2y' = y^2 + 2xyy' Solving for y', y' = (2x+y-y^2)/(2xy-x+3y^2).. More from Yahoo Answers
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