Factor out the term = sin x + cos x. Now, we can isolate
, which can be simplified to
30.
We take the derivative of each term with respect to x: = .
Next, because = 1, we can eliminate that term to get = .
Next, don’t simplify. Plug in (1, 1) for x and y, = , which
simplifies to .
Finally, we can solve for .
SOLUTIONS TO PRACTICE PROBLEM SET 8
- y − 2 = 5(x − 1)
Remember that the equation of a line through a point (x 1 , y 1 ) with slope m is y − y 1 = m(x − x 1 ).
We find the y-coordinate by plugging x = 1 into the equation y = 3x^2 − x, and we find the slope
by plugging x = 1 into the derivative of the equation.
First, we find the y-coordinate, y 1 : y = 3(1)^2 − 1 = 2. This means that the line passes through
the point (1, 2).
Next, we take the derivative: = 6x − 1. Now, we can find the slope, m: = 6(1) − 1 =
- Finally, we plug in the point (1, 2) and the slope m = 5 to get the equation of the tangent line:
y − 2 = 5(x − 1).
- y − 18 = 24(x − 3)
Remember that the equation of a line through a point (x 1 , y 1 ) with slope m is y − y 1 = (x − x 1 ).
We find the y-coordinate by plugging x = 3 into the equation y = x^3 − 3x, and we find the slope
