find dy/dx when y=(cosx + 11x)/(e^x-tanx)?

Answer 1

Unfortunately, not many terms cancel:

dy/dx = {-sinx*(e^x) + 11*(e^x) + tanx*sinx - 11*tanx - cosx*(e^x) + cosx*((secx)^2) - 11x*(e^x) + 11x*((secx)^2)} / {((e^x) - tanx)^2}

Whew, hope that helps

Answer 2

y = (cosx + 11x)/(e^x - tanx)

f'(x) = (f'(x)g(x) - f(x)g(x)')/(g(x))^2

f(x) = cosx + 11x
g(x) = e^x - tanx

f'(x) = (((cosx + 11x)'(e^x - tanx)) - ((cosx + 11x)(e^x - tanx)))/((e^x - tanx)^2)

f'(x) = ((sinx + 11)(e^x - tanx)) - ((cosx + 11x)(e^x - secx^2)))/((e^x - tanx)^2)

f'(x) = ((11xsec(x)^2) + secx - sinxtanx + 11tanx + 11e^x - 11xe^x - (e^x)cosx + (e^x)sinx)/((e^x - tanx)^2)

Answer 3

2?

Answer 4

Use the Quotient Rule: (Denom*dNum-Num*dDenom)/Denom^2

In other words:
[(e^x-tan x)*d(cos x + 11x) - (cos x +11x)*d(e^x - tan x)]/(e^x-tan x)^2

Where:
d(cos x -11x) = 11-sin x
d(e^x-tan x) = e^x - 1/(cos x)^2

And I'm pretty sure you can do the multiplication....