implicit differentiation

[dukefan1342]

I did both these problems on homework a few days back but theres a test tomorrow and i still dont completely understand how to do these. if anyone could help explain them that would be great. For both i had to find y' using implicit differentiation
1. x²y² + x sin(y) = 1

I got this down to (-8x^7(x+y)+9y^2-x^8)/ (x^8-2y(9x-y)-y^2)

2. x⁸(x + y) = y²(9x − y)
I am completely stuck on this one

 

[Jason76]

Implicitly differentiate with respect to $\displaystyle x$ - This means solve for $\displaystyle y'$.

Remember to differentiate everything, but only the differentiated y stuff, has y' connected to it.

1.

$\displaystyle x^{2}y^{2} + x\sin(y) = 1$

$\displaystyle [(2x)(y^{2}) + (2yy')(x^{2})] + [(1)(\sin(y)) + (\cos(y)y')(x)] = 0$ Do product rule for terms on the left of the equation. On the right the derivative of a constant is $\displaystyle 0$.

$\displaystyle (2yy')(x^{2}) + (\cos(y)y')(x) + (2x)(y^{2}) + (1)(\sin(y)) = 0$ - Group y' containing stuff together.

$\displaystyle (2yy')(x^{2}) + (\cos(y)y')(x)] = - (2x)(y^{2}) - (1)(\sin(y))$ - Move terms that don't have y' connected to them on the other side of the equation.

$\displaystyle y'[(2y)(x^{2}) + (\cos(y))(x)] = -(2x)(y^{2}) - (1)(\sin(y)) = 0$ - Factor out y'.

$\displaystyle y' = \dfrac{- (2x)(y^{2}) - (1)(\sin(y)) }{ (2y)(x^{2}) + (\cos(y))(x)}$ - Isolate y'

$\displaystyle y' = \dfrac{- (2x)(y^{2}) - (\sin(y)) }{ (2y)(x^{2}) + (\cos(y))(x)}$ - Sometimes the answer doesn't show the 1 values.

The answer might be written like this, which is the same thing:

$\displaystyle y' = \dfrac{- (2x)(y^{2}) - (\sin(y)) }{ (2x^{2})(y) + (x)( \cos(y))}$

2.

$\displaystyle x^{8}(x + y) = y^{2}(9x - y)$

$\displaystyle (8x^{7})(x + y) + (1 + (1)y')(x^{8}) = (2yy')(9x - y) + (9 - (1)y')(y^{2})$ - Use the product rule for terms, some of which are contained in parenthesis.

$\displaystyle (1 + (1)y')(x^{8}) - (2yy')(9x - y) - (9 - (1)y')(y^{2}) = -(8x^{7})(x + y)$ Group y' terms on the left. The non y' stuff on the right.

$\displaystyle y'[ (1 + (1))(x^{8}) - (2y)(9x - y) - (9 - (1))(y^{2})] = -(8x^{7})(x + y)$ Factor out y'

$\displaystyle y' = \dfrac{ -(8x^{7})(x + y) }{ (1 + (1))(x^{8}) - (2y)(9x - y) - (9 - (1))(y^{2}) }$ Isolate y'

$\displaystyle y' = \dfrac{ -(8x^{7})(x + y) }{ (2x^{8}) - 18xy + 2y) - 8(y^{2}) }$ Might be simplifed like this.

 

[HallsofIvy]

I did both these problems on homework a few days back but theres a test tomorrow and i still dont completely understand how to do these. if anyone could help explain them that would be great. For both i had to find y' using implicit differentiation
1. x²y² + x sin(y) = 1

I got this down to (-8x^7(x+y)+9y^2-x^8)/ (x^8-2y(9x-y)-y^2)

2. x⁸(x + y) = y²(9x − y)
I am completely stuck on this one

??? You have (1) and (2) reversed! What you give, in (1) as the derivative of $\displaystyle x^2y+ x sin(y)= 1$ is the derivative of (2) $\displaystyle x^8(x+ y)+ y= y^2(9x- y)$!

To differentiate $\displaystyle x^2y^2+ x sin(y)= 1$, by the product rule, $\displaystyle (x^2y^2)'= (x^2)'y^2+ x^2(y^2)'= (2xx')y^2+ x^2(2yy')$. Since we are differentiating with respect to x, x'= 1 so $\displaystyle (x^2yy^2)'= 2xy^2+ 2x^2yy'$. Also by the product rule, (x sin(y))'= (x') sin(y)+ x(sin(y))'. I presume you know that (sin(y))'= cos(y) y' so this is (1)sin(y)+ x cos(y) y'.

The derivative of left side of the equation is $\displaystyle 2xy^2+ 2x^2yy'+ sin(y)+ x cos(y)y'$. The right side is "1", a constant, so its derivative if 0.
Solve $\displaystyle 2xy^2+ 2x^2yy'+ sin(y)+ x cos(y)y'= 0$ for y'.