Help with Derivative and Tangent Line Problem

[cummings15]

First Problem

Homework Statement

Find the derivative of x^6+y^6=18xy

Homework Equations

Find derivative

The Attempt at a Solution

6x^5+6y^6=18*(dy/dx)

Second Problem

Homework Statement

Verify that (1,0) is on the following curve and find the tangent line and normal line to the curve point. y=pisin(pix-y)

Homework Equations

Tangent line derivative

The Attempt at a Solution

tangent line = pi^2cos(pix-y)

 

[Quinzio]

Find the derivative of x^6+y^6=18xy

6x^5+6y^6=18*(dy/dx)

?

No really...

This is an implicit function, basically it's a curve on the xy plane.

I'd do like this
$$f_x = 6x^5-18y$$
$$f_y = 6y^5-18x$$

$$\frac{dy}{dx} = -\frac{6x^5-18y}{6y^5-18x}$$

 

[HallsofIvy]

As Quinzio said, you don't differentiate an equation, you differentiate a function.

Quinzio assumed you really meant "differentiate y with respect to x where y is defined by this equation". Given that same equation, it could as well have been "differentiate x with respect to y."

An equivalent way to do it is this:
x^6+ y^6= 18xy
Differentiating both sides, with respect to x,
6x^5+ 6y^5 y'= 18y+ 18xy' where I have written the derivative of y itself, with respect to x, as y' and have used the chain rule to differentiate y^6 with respect to x.
Then we have (6y^5- 18x)y'= 18y- 6x^5 so that y'= (18y- 6x^5)/(6y^5- 18x) which is exactly what Quinzio has.

You have several mistakes. You have "6y^6" rather than "6y^5" which may be a typo. More importantly, you did multiply by dy/dx and so did not solve for dy/dx.

As for the tangent problem, you did not do the first part, "verify that (1, 0) is on the curve" but I assume you were able to do that and are just not asking about it.

Far, far, far more important is that your answer is not even the equation of a line! You do understand, I hope, that the "tangent line" to any curve is, first, a line?

With $y= \pi sin(\pi x- y)$ you need to differentiate y with respect to x. Again, that means "implicit differentiation" because y is an "implicit" function of x (i.e. is not just "y= f(x)"). Differentiating both sides with respect to x, $y'= \pi sin(\pi x- y)(1- y')$. I have used the chain rule to do that: writing $u= \pi x- y$, gives $y= \pi sin(u)$ so that, differentiating with respect to x, $y'= \pi cos(u) (du/dx)$ and, since $u= \pix- y$, du/dx= \pi- y'[/itex].

Solve that for y'. And, remember that the derivative is not "the tangent line". It is only the slope of the tangent line. Once you have found y', as a function of x and y, evaluate it at x= 1, y= 0 to find the slope of the tangent line, a number.

You can write the equation of the line in "slope, intercept" form, y= mx+ b, or "slope, point" form, $y= m(x- x_0)+ y_0$.

 

[cummings15]

so for the function y=pisin(pix-y)

y prime = picos(pix-y)*pi

which equals pi^2cos(pix-y)