when y=5tanx-15cos x, find dy/dx, at x="pi"/4?

Question

stuck on this question. thanks

Answer 1

dy/dx=5sec^2x+15sinx
when x=pi/4
5*2+15rt2/2
=10+7.5 rt2

Answer 2

y = 5tan(x) - 15cos(x)

To find dy/dx at x = pi/4, all you have to is take the derivative, and then upon doing that, plug in x = pi/4 afterward. What this represents is solving for the slope of the tangent line of the graph at x = pi/4.

Solving for dy/dx, we take the derivative.

dy/dx = 5sec^2(x) - 15(-sin(x))
dy/dx = 5sec^2(x) + 15sin(x)

Now, we plug in x = pi/4. Let's call the new form "m" to represent the slope of the tangent line at x = pi/4.

m = 5sec^2(pi/4) + 15sin(pi/4)

Note that secant squared pi/4 is equal to all of secant pi/4 squared.

m = 5[sec(pi/4)]^2 + 15sin(pi/4)

Now we solve normally. It's not immediately obvious what secant of pi/4 is, so we'll change it to cosine.

m = 5[1/[cos(pi/4)]^2 + 15sin(pi/4)
m = 5[1/(sqrt(2)/2)]^2 + 15(sqrt(2)/2)
m = 5[1/(2/4)] + 15sqrt(2)/2
m = 5[4/2] + 15sqrt(2)/2
m = 20/2 + 15sqrt(2)/2
m = [20 + 15sqrt(2)]/2

If we want a neater answer,

m = (5/2) [4 + 3sqrt(2)]

Answer 3

When y=5tanx-15cos x, find dy/dx, at x="pi"/4?
dy/dx = 5sec^2x +15sinx
= 10 +10.6 = 20.6 when x = pi/4 = 45 degrees.

I used 1/cos^x for sec^2x.

Answer 4

y = 5tan x - 15cos x

Find dy/dx at x = π/4

dy/dx = 5sec² x + 15sin x
= 5sec²(π/4) + 15sin(π/4)
= 5(√2)² + 15(1/√2) = 5*2 + 15(1/√2)
= 10 + 15/√2

Answer 5

y'=5sec^2(x)+15sinx(x)