Help with Derivative and Tangent Line Problem

In summary: So the slope of the tangent line is pi^2cos(pi-0)= -pi^2Using the point-slope form, the equation of the tangent line is y= -pi^2(x-1)+0= -pi^2x+pi^2The normal line will have a slope that is the negative reciprocal of the slope of the tangent line, so the slope of the normal line is 1/pi^2. Using the point-slope form again, the equation of the normal line is y= (1/pi^2)(x-1)+0= x/pi^2 - 1/pi^2In summary, the derivative of x^6+y
  • #1
cummings15
17
0
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)
 
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  • #2
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
[tex] f_x = 6x^5-18y[/tex]
[tex] f_y = 6y^5-18x[/tex]

[tex]\frac{dy}{dx} = -\frac{6x^5-18y}{6y^5-18x}[/tex]
 
  • #3
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 [itex]y= \pi sin(\pi x- y)[/itex] 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, [itex]y'= \pi sin(\pi x- y)(1- y')[/itex]. I have used the chain rule to do that: writing [itex]u= \pi x- y[/itex], gives [itex]y= \pi sin(u)[/itex] so that, differentiating with respect to x, [itex]y'= \pi cos(u) (du/dx)[/itex] and, since [itex]u= \pix- y[/itex], 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, [itex]y= m(x- x_0)+ y_0[/itex].
 
  • #4
so for the function y=pisin(pix-y)

y prime = picos(pix-y)*pi

which equals pi^2cos(pix-y)
 

1. What is a derivative?

A derivative is a mathematical concept that represents the rate of change of a function at a specific point. It gives the slope of a tangent line to the function at that point.

2. How do I find the derivative of a function?

To find the derivative of a function, you can use the rules of differentiation such as the power rule, product rule, quotient rule, and chain rule. These rules help you find the derivative of different types of functions.

3. What is a tangent line?

A tangent line is a line that touches a curve at only one point, and it has the same slope as the curve at that point. It represents the instantaneous rate of change of the curve at that point.

4. How do I use derivatives to solve real-life problems?

Derivatives can be used to solve a variety of real-life problems such as finding the maximum or minimum value of a function, calculating the velocity and acceleration of an object, and optimizing the production or cost of a product.

5. What are some common mistakes to avoid when solving derivative and tangent line problems?

Some common mistakes to avoid when solving derivative and tangent line problems include using incorrect differentiation rules, forgetting to simplify the final answer, and not correctly identifying the point at which the tangent line is drawn. It is also important to carefully read the problem and understand what is being asked before attempting to solve it.

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