4
For f(x) + g(x), the derivative d/dx[f(x) + g(x)] = f'(x) + g'(x). Essentially, this just means that since it's addition, you can take the derivative of each part.d/dx(x - 2cosx) =* d/dx is a way to indicate you're taking the derivatived/dx(x) + d/dx(-2cosx)* take the derivative of each part, distributing(1) + (-2*-sinx)* f(x) = x, f'(x) = 1 & g(x) = -2cosx, g'(x) = +2sinx= 1 + 2sinx
d/dx 2 cos x = -2 sin x
cos2x/cosx = 2cosx - 1/cosx
if y=1/4x and x=,find y
Assuming that you want to find y when x is 40, it is 200.
Oh, finding the x and y intercepts is like finding little treasures in your painting. To find the x-intercept, you set y to zero and solve for x. To find the y-intercept, you set x to zero and solve for y. Remember, there are no mistakes in mathematics, just happy little accidents.
To find x as a percentage of y, calculate 100*x/y.
solve the x and you find the y
d/dx (2cos(x)sin⁻¹(x)) right, 2cos(x)sin⁻¹(x) is a product, so we'll differentiate using the product rule: d/dx(uv) = u d/dx(v) + v d/dx(u) u = 2cos(x) v = sin⁻¹(x) d/dx(u) = -2sin(x) to find d/dx(sin⁻¹(x)) we'll set y=sin⁻¹(x) x=sin(y) dx/dy = cos(y) dy/dx = 1/(cos(y)) cos²(y) + sin²(y)=1, dy/dx = 1/sqrt(1-sin²y) = 1/sqrt(1-x²) [x=sin(y)] so plugging all this into our product rule, d/dx (2cos(x)sin⁻¹(x)) = 2cos(x)/sqrt(1-x²) - 2sin(x)sin⁻¹(x).
You look for the value of 0 in the y column, and find out what x has to be for y=0. This value of x is you x-axis intercept. (Reverse "x" and "y" in the above description to find the y-intercept, if there is one).
to find the y-intercept you plug in your x and y values in to the equation of y=mx+b. b is the y intercept and m is the slope. To find the x-intercept, set y = 0, and find value of x that satisfies the equation. If it is a line in the form y=mx+b, then the x-intercept will be at x= -b/m