Question

Use the quotient rule and the derivative of the cosine function to prove that $\frac{d}{dx}(secx)=secx.\tan x$

Short answer

The derivative $\frac{d}{dx}(secx)=secx.\tan x$ has been proved.

Step-by-step solution

Step 1. Given Information.

The function:

$\frac{d}{dx}(secx)=secx.\tan x$

Step 2. Quotient rule.

$\left(\frac{f}{g}\right)\text{'}\left(x\right)=\frac{f\text{'}\left(x\right)g\left(x\right)-g\text{'}\left(x\right)f\left(x\right)}{(g{\left(x\right))}^2}$

Step 3. Use the quotient rule on the function.

$$\begin{gathered} f\text{'}\left(secx\right)=f\text{'}\left(\frac{1}{\cos x}\right) \\ =\frac{\left(1\right)\text{'}\left(\cos x\right)-1\left(\cos x\right)\text{'}}{{\left(\cos x\right)}^2} \\ =\frac{-(-\sin x)}{{\cos }^{2}x} \\ =\frac{1}{\cos x}.\frac{\sin x}{\cos x} \\ =secx.\tan x \end{gathered}$$