Question

Find dy/dx by implicit differentiation.
4x2 + 5xy - y2 = 6

Answer 1

\[4x^2+5xy-y^2=6\]

Answer 2

8x + 5y + 5xdy/dx - 2ydy/dx = 0
dy/dx = (8x+5y)/(2y-5x)

Answer 3

please show me step by step so that i can follow up what i am doing

Answer 4

isolate the dy/dx

Answer 5

aint nuthing different about implicit than normal;

Answer 6

4x^2 becomes 8x naturally
5xy we need to use the product rule so 5y + 5xy' (taking dy/dx y becomes (1)y' from chain rule)
-y^2 comes to be -2y * y' from chain rule

Answer 7

\[4x^2 + 5xy - y^2 = 6 \]
\[\frac{d(4x^2)}{dx}+\frac{d(5xy)}{dx}-\frac{y^2)}{dx}=\frac{d(6)}{dx}\]

Answer 8

you need to keep in mind that you are taking the derivative with respect to x and not to y

Answer 9

\[\frac{dx}{dx}8x+\frac{dx}{dx}5y+5x\frac{dy}{dx}-2y\frac{dy}{dx}=\frac{dx}{dx}0\]

Answer 10

i know kanade but its gets really hard to understand some of the function

Answer 11

\[{dx\over dx}=1; \frac{dy}{dx}=y'\]
\[8x +5y +5x.y' -2y.y' = 0\]

Answer 12

ok now it looks good..but let me see if i got the same answer

Answer 13

\[5x.y' -2y.y' = -8x-5y\]
\[y'(5x -2y)=-8x-5y\]
\[y' = \frac{-8x-5y}{5x-2y}\]

Answer 14

when you take dy/dx y^2 (with respect to x)
y^2 is treated like a function but you dont know what that function is

so like if you take cos^2(SOME FUNCTION)
you would do 2(-sin(SOME FUNCTION))d/dx(SOME FUNCTION)

Answer 15

you are used to the "baby" derivatives.

\[y=x^2\]
\[\frac{d(y)}{dx}=\frac{d(x^2)}{dx}\]
\[1*\frac{dy}{dx}=\frac{dx}{dx}*2x\]
\[y' = 2x\]
but that is the same process you use in 'implicit'
\[y^2 = x^2\]
\[\frac{d(y^2)}{dx}=\frac{d(x^2)}{dx}\]
\[2y\frac{dy}{dx}=\frac{dx}{dx}2x\]
\[2y.y' = 2x\]
\[y' = \frac{2x}{2y}=\frac{x}{y}\]

Answer 16

ok cool thanks soo much