Question

Find the derivative of lan (tanx+secx)

Answer 1

To find the derivative of ln(tanx+secx), we apply the chain rule. The derivative is (1/(tanx+secx))*(sec^2(x)+tan(x)*sec(x)).

Step-by-step explanation:

To find the derivative of ln(tanx+secx), we can use the chain rule. Let's denote the function as f(x) = ln(tanx+secx). The chain rule states that if we have a composite function f(g(x)), the derivative is given by f'(g(x)) * g'(x).

Applying the chain rule, we have:

f'(x) = (1/(tanx+secx))*(sec^2(x)+tan(x)*sec(x))

Apply the chain rule: [ \frac{d}{dx}[\ln(\tan x + \sec x)] = \frac{1}{\tan x + \sec x} \cdot (\sec^2 x + \sec x \tan x) ].

Therefore, the derivative of ln(tanx+secx) is (1/(tanx+secx))*(sec^2(x)+tan(x)*sec(x)).

So, the derivative is: [ \frac{\sec^2 x + \sec x \tan x}{\tan x + \sec x} ].