What is the derivative of rπx+q*sin(w)

What is the derivative of rπx+q*sin(w)?
w, q, r are constants. Solve with respect to x.

d (r π x+q*sin(w)) / dx =
r π

since +qsinw is a constant not a coefficient of x.

Step 1, split this into two derivatives:

dy/dx rπx+q Since the derivative of a constant is zero, you only need worry about the rπx whose derivative is simply rπ.

dy/dx q*sin(w) You need to use the product rule for this one. f[x] * g'[x] + f'[x] * g[x]

let q be f[x], then f'[x] will be 0 since the derivative of a constant is zero.

let sin(w) be g[x], then g'[x] will need the chain rule, derivative of the outside, leave the inside, times the derivative of the inside. So, g'[x] would be Cos[w]*0 = 0.

Now we need to recombine :

f[x] * g'[x] + f'[x] * g[x]

q * 0 + 0 * sin(w) = 0.

Now we have to recombine the two derivatives from the beginning:

All we are left with is rπ.

dy/dx = x + [ q * cos(w) * w]

You have to use the combination of product rule and quotient rule for the second term in this expression.

Hope this helps

rπ