find dy/dx
y=ln (secx + tanx)
Let u= secx + tan x
dy/dx= 1/u * du/dx
now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it.
Use the chain rule. Let y(u) = ln u
u(x) = sec x + tan x
dy/dx = dy/du*du/dx
dy/du = 1/u = 1/(sec x + tan x)
dy/dx = sec x tan x + sec^2 x
= sec x (sec x + tan x)
dy/dx = sec x
(sec x + tan x)
So, the derivative dy/dx of y = ln(sec x + tan x) is:
dy/dx = (sec x)(sec x + tan x) / (sec x + tan x)
dy/dx = sec x.
dy/dx = sec x
To find dy/dx, we start by letting u = secx + tanx. Now we have the function y = ln(u), and we want to find dy/dx.
Using the chain rule, we have dy/dx = dy/du * du/dx.
First, we need to find dy/du. We know that the derivative of ln(u) is 1/u. Therefore, dy/du = 1/u.
Next, we need to find du/dx. Since u = secx + tanx, we can find du/dx using the derivative rules for secant and tangent functions. The derivative of secx is secx * tanx, and the derivative of tanx is sec^2x. Therefore, du/dx = secx * tanx + sec^2x.
Now, we can substitute the values we found into our equation for dy/dx.
dy/dx = dy/du * du/dx
= (1/u) * (secx * tanx + sec^2x)
Substituting u = secx + tanx, we have:
dy/dx = (1/(secx + tanx)) * (secx * tanx + sec^2x)
Simplifying the expression, we get:
dy/dx = secx
So, the derivative of y = ln(secx + tanx) with respect to x is dy/dx = secx.