# What would m be?

equation of the tangent line to the graph of y=xcos(3x) at x=π is given by y=mx+bfor

### 4 Answers

- PhilipLv 61 month ago
y = xcos(3x). y' = cos(3x) + x[-sin(3x)](3) = cos(3x) -3xsin(3x). At x = pi, y' =;

cos(3pi) - 3pi*sin(3pi) = -1. Also, at x = pi, y = pi*cos(3pi) = -pi.;

Eqn of tangent line L to y at x = pi is y = mx + b where m is the slope of tangent line L and b is the y-intercept of L. At x = pi, m = -1. Anonymous only wanted to

know the value of m, not the eqn of the tangent line {y = -x}.

- Engr. RonaldLv 71 month ago
Differentiate the function

y = xcos(3x)

dy/dx = cos(3x) d/dx 1 + x d/dx - 3sin(3x)

dy/dx = cos(3x) - 3xsin(3x)

when x = π

m = dy/dx = cos(3π) - 3πsin(3π)= - 1

the slope is m = - 1

solving for y

y = πcos(3π) = -π

solving its y - intercept from slope intercept form

y = mx + b

- π = - π + b

- π + π = b

b = 0

The equation of the tangent line is:

y = - x Answer//

- llafferLv 71 month ago
Need the derivative of:

y = x cos(3x)

Product rule:

u = x and v = cos(3x) so we have y = uv

y' = u'v + v'u

so we need the derivatives of u and v. u is easy, v needs the chain rule:

u' = 1

v = cos(3x)

setting: w = 3x and v = cos(w)

dv/dx = dv/dw * dw/dx

Now we get the derivatives of w and v:

w = 3x and v = cos(w)

dw/dx = 3 and dv/dw = -sin(w)

Substitute 3x back in for w to get:

dw/dx = 3 and dv/dw = -sin(3x)

Now we have what we need to complete the chain rule:

dv/dx = dv/dw * dw/dx

dv/dx = -sin(3x) * 3

v' = -3sin(3x)

Now we have what we need to complete the product rule. Summarizing what we know:

u = x and v = cos(3x)

u' = 1 and v' = -3sin(3x)

y' = u'v + v'u

y' = 1cos(3x) + (-3)sin(3x)x

y' = cos(3x) - 3x sin(3x)

Now that we have this, the slope at x = π can be found by solving for y'(π):

y' = cos(3π) - 3π sin(3π)

y' = -1 - 3π(0)

y' = -1 - 0

y' = -1

The slope is -1.

We are given x = π, so now we need to find "y" at this point:

y = x cos(3x)

y = π cos(3π)

y = π(-1)

y = -π

Now that we have the slope, x, and y, we can solve for the intercept:

-π = (-1)π + b

-π = -π + b

0 = b

The tangent line at x = π is:

y = -x

- AshLv 71 month ago
y=xcos(3x)

y' = cos(3x) + x (-sin(3x) * 3)

y' = cos(3x) - 3x sin(3x)

y' = slope of tangent

At x = π,

slope of tangent = cos(3π) - 3π sin(3π)

slope of tangent = -1 - 3π (0)

slope of tangent = -1

m = -1

If you want to find the equation of line then find the y-coordinate for point with x=π

y=πcos(3π)

y = -π

tangent point is (π,-π)

y-(-π) = -1(x-π)

y+π = -x+π

y = -x ← equation of tangent