Derivative of the Secant Function

The first derivative of the secant function is given by $$ D[\sec \ x] = \frac{\sin x}{\cos^2 x} = \sec \ x \cdot \tan \ x. $$

Proof

The secant function is defined as the reciprocal of the cosine function:

$$ \sec x = \frac{1}{\cos x}. $$

Therefore, we can write the derivative of the secant function as:

$$ D[\sec \ x] = D\left[ \frac{1}{\cos x} \right]. $$

Applying the quotient rule, we get:

$$ D[\sec \ x] = \frac{D[1] \cdot \cos x - 1 \cdot D[\cos x]}{\cos^2 x}. $$

Since the derivative of a constant is zero, $D[1] = 0$.

Substituting these results, we obtain:

$$ D[\sec \ x] = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x}. $$

Which simplifies to:

$$ D[\sec \ x] = \frac{\sin x}{\cos^2 x}. $$

We can further rewrite this expression as:

$$ D[\sec \ x] = \frac{\sin x}{\cos x} \cdot \frac{1}{\cos x}. $$

In trigonometry, the ratio $\sin x / \cos x$ corresponds to the tangent function:

$$ D[\sec \ x] = \tan x \cdot \frac{1}{\cos x}. $$

Since the reciprocal of cosine is the secant function, we have:

$$ D[\sec \ x] = \tan x \cdot \sec x. $$

Thus, we have established the formula for the derivative of the secant function.

And so on.