Where Does the 3x Come From in e^(sec 3x)?

  • Thread starter Thread starter powp
  • Start date Start date
  • Tags Tags
    Explain
powp
Messages
91
Reaction score
0
can someone explain where the 3x comes from in the below example

y = e^(sec 3x)

dy/dx = e^(sec 3x) * d/dx(sec 3x)
= e^(sec 3x) * sec 3x tan 3x d/dx(3x)
^
where does this come from??
 
Physics news on Phys.org
Apply the chain rule to sec(3x).
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top