# What is the slope of the line normal to the tangent line of #f(x) = sec^2x-xcos(x-pi/4) # at # x= (15pi)/8 #?

Interactive graph

For the 2nd term, we'll need to use a product rule. So:

Now, we put everything together:

Watch your signs.

However, what we want is not the line tangent to f(x), but the line normal to it. To get this, we just take the negative reciprocal of the slope above.

Now, we just fit everything into point slope form:

#y = m(x-x_0) + y_0

Take a look at this interactive graph to see what this looks like!

Hope that helped :)

By signing up, you agree to our Terms of Service and Privacy Policy

To find the slope of the line normal to the tangent line of ( f(x) = \sec^2(x) - x\cos(x - \frac{\pi}{4}) ) at ( x = \frac{15\pi}{8} ), we need to first find the slope of the tangent line at that point and then find the negative reciprocal of that slope to get the slope of the normal line.

The slope of the tangent line at ( x = \frac{15\pi}{8} ) can be found by taking the derivative of ( f(x) ) and evaluating it at that point.

( f'(x) = 2\sec^2(x)\tan(x) + \cos(x - \frac{\pi}{4}) + x\sin(x - \frac{\pi}{4}) )

Evaluate ( f'(\frac{15\pi}{8}) ) to get the slope of the tangent line at ( x = \frac{15\pi}{8} ).

Then, take the negative reciprocal of this slope to find the slope of the normal line.

By signing up, you agree to our Terms of Service and Privacy Policy

- The equation of a circle is #(x- 3)^2 + (y +2)^2= 25#. The point (8 -2) is on the circle What is the equation of the line that is tangent to the circle at (8,-2)?
- How do you find the average rate of change of #g(x)=1/(x-2)# over [0,3]?
- How do you find the average rate of change of #f(x)=cot x# over the interval [pi/6, pi/2]?
- How do you use the limit definition to find the derivative of #y=sqrt(-4x-4)#?
- How to find instantaneous rate of change for #f(x) = 3x^2+4x# at (1, 7)?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7