-find the equation of the tangent line to the curve y=5xcosx at the point (pi,-5pi)

-the equation of this tangent line can be written in the form y=mx+b where
m=
and b=
-what is the answer to m and b?

Given

(x0,y0)=(π, -5π)

We have established:
m
=f'(π)
=5(cos(π)-π*sin(π))
=5(-1 - π*0)
=-5

The equation of a line of slope m passing through (x0,y0) is:
(y-y0)=m(x-x0)
(y-(-5π))=(-5)(x-π)
Simplify to get the required equation.
b=0 by coincidence.

See graph:
http://img10.imageshack.us/i/1298601541.png/

Check my work.

you did not follow MathMate's suggestion

f'(x) = 5(cosx - xsinx)
f'(π) = 5(cosπ - πsinπ)
= 5(-1 - π(0)) = -5 , so m = -5

equation:
y = -5x + b , but (π,-5π) lies on it, so
-5π = -5(π) + b
b = 0

equation : y = -5x

or using MathMate's suggestion:

y + 5π = -5(x-π)
y + 4π = -5x + 5π
y = 5x

ok how about this one:

f(x)=12x/sinx+cosx
find f'(-pie)
this is what i did:
12(-pi)/(0+(-1))
= 12pi
is that right what am i doing wrong and what would be the answer

Well, you know what they say about tangents, they're the line that just can't help but touch a curve at one point. Now, let's find this tangent for y = 5xcos(x) at the point (π, -5π).

To find the equation of the tangent line, we need to find the slope (m) and the y-intercept (b). So let's get cracking.

First, we need to find the derivative of y with respect to x. Taking the derivative of 5xcos(x) is no piece of cake, but luckily it's not as scary as a clown at a birthday party. The derivative of 5xcos(x) is 5cos(x) - 5xsin(x).

Now, to find the slope (m), we need to evaluate this derivative at the x-coordinate of the given point, which is π. So, substitute π into the derivative: m = 5cos(π) - 5πsin(π).

Now, let's simplify this. Cosine of π is -1, and sine of π is 0, so m = 5(-1) - 5π(0) = -5.

Great, we have found the slope (m) of the tangent line. Now, onto finding the y-intercept (b). To do that, we can use the point-slope formula:

y - y1 = m(x - x1).

Substituting the values of (x1, y1) as (π, -5π) and m as -5, we get: y - (-5π) = -5(x - π).

Simplifying this will give us the equation for the tangent line: y = -5x - 5π + 5π.

And you know what? We can simplify that even further! The -5π and +5π will cancel each other out, leaving us with y = -5x as the equation of the tangent line.

So the answer is: m = -5 and b = 0. Yes, you heard that right - b = zero, nothing, zilch. That means the y-intercept is right there at the origin. No clowning around!

So, the equation of the tangent line is y = -5x. Happy math-ing!

To find the equation of the tangent line to the curve y=5xcosx at the point (pi, -5pi), we need to find the slope of the tangent line and the y-intercept.

Step 1: Find the slope (m) of the tangent line
The slope of the tangent line can be found by taking the derivative of the given function with respect to x.

Since y = 5xcosx, we can find the derivative using the product rule:

dy/dx = 5x * d(cosx)/dx + cosx * d(5x)/dx

Let's evaluate each term separately:
1. d(cosx)/dx: The derivative of cosine function is -sinx. So, d(cosx)/dx = -sinx.
2. d(5x)/dx: The derivative of 5x is simply 5.

Now, substituting these results back into the derivative equation:

dy/dx = 5x*(-sinx) + cosx*5
= -5xsinx + 5cosx

To find the slope of the tangent line at the point (pi, -5pi), substitute x = pi into the derivative equation:

m = -5(pi)sin(pi) + 5cos(pi)
= -5pi(0) + 5(-1)
= -5(-1)
= 5

So, the slope (m) of the tangent line is 5.

Step 2: Find the y-intercept (b) of the tangent line
To find the y-intercept, we can substitute the coordinates of the given point (pi, -5pi) into the equation y = mx + b and solve for b.

Using the coordinates (pi, -5pi) and the slope m = 5:

-5pi = 5(pi) + b

Simplifying the equation:

-5pi = 5pi + b
b = -5pi - 5pi
b = -10pi

So, the y-intercept (b) of the tangent line is -10pi.

Therefore, the equation of the tangent line can be written as:

y = 5x - 10pi

Hence, m = 5 and b = -10pi.

m=pi

and b= y-pi=pi(x+5pi)
y-pi=pi(x)+5pi(pi)
m=pie but what does b= though

f(x)=5xcos(x)

f'(x)=5cos(x)-5xsin(x)
=5(cos(x)-xsin(x))
m=f'(π)
and the equation of the tangent passing through the point (x0,y0)=(π, -5π) is:
(y-y0)=m(x-x0)
Substitute values and simplify to get the equation of the line.