# Derivative of x sec(x) with Proofs and Graphs

The function x multiplied by the secant of x is a product of the monomial function x and the trigonometric function of secant. The derivative of x multiplied by the secant of x is equal to the quantity of x multiplied by the secant and tangent of x and then added to the secant of x, xsec(x)tan(x) + tan(x). This derivative can be proved using the product rule, implicit differentiation, quotient rule, trigonometric secant function, or the inverse secant function.

In this article, we will discuss how to derive the product of two functions, the monomial x and the trigonometric function secant. We will cover brief fundamentals, its formula, a graph comparison of secant and its derivative, proofs, methods to derive, and an example.

##### GEOMETRY

Relevant for

Learning how to find the derivative of x sec(x).

See proofs

##### GEOMETRY

Relevant for

Learning how to find the derivative of x sec(x).

See proofs

## Proofs of the Derivative of x sec(x)

The proofs of the derivative of $$x\sec{(x)}$$ are listed below. These proofs can also be used as methods of deriving this function.

### Proof of the derivative of x sec(x) using the Product Rule formula

You can review the product rule formula by looking at this article: Product Rule of derivatives. You may also look at the proof of the trigonometric secant derivative in this article: Derivative of Secant, sec(x).

Suppose we are asked to derive the function

$$f(x) = x\sec{(x)}$$

We may deduce which two functions are being multiplied. It is a product of a monomial and the trigonometric secant of angle x in this instance. With u as the first multiplicand/term, we have

$$u = x$$

and with v as the second multiplicand/term, we have

$$v = \sec{(x)}$$

Keep in mind that the product formula for derivative is

$$\frac{d}{dx} uv = u \frac{d}{dx} v + v \frac{d}{dx} u$$

In other words, the function u times v is derived by multiplying u by the derivative of v and then adding it to v multiplied by the derivative of u. When we apply this formula to our given function, we get

$$\frac{d}{dx} uv = (x) \frac{d}{dx} (\sec{(x)}) + (\sec{(x)}) \frac{d}{dx} (x)$$

Evaluating, we have the derivative of u using the power rule and the derivative of v using the derivative of secant.

$$\frac{d}{dx} uv = (x) \cdot (\sec{(x)}\tan{(x)}) + (\sec{(x)}) \cdot (1)$$

Simplifying, we have

$$\frac{d}{dx} uv = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

Resulting to the derivative formula of $$x\sec{(x)}$$

$$\frac{d}{dx} x\sec{(x)} = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

### Proof of the derivative of xsec(x) using implicit differentiation through quotient rule and trigonometric secant function

Having the equation

$$y = x\sec{(x)}$$

By cross multiplying the first multiplicand of the right-hand side to the left-hand side of the equation, we get

$$\frac{y}{x} = \sec{(x)}$$

In terms of x, evaluate the implicit differentiation. On the left hand side, we will utilize the quotient rule, and on the right, we’ll use the derivative of secant.

$$\frac{d}{dx} \left( \frac{y}{x} \right) = \frac{d}{dx} \sec{(x)}$$

Keep in mind that the quotient rule formula is

$$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u \frac{d}{dx} v – v \frac{d}{dx} u}{v^2}$$

Applying this to the derivative of the left hand side of the equation, we have

$$\frac{x \frac{dy}{dx} – y}{x^2} = \frac{d}{dx} \sec{(x)}$$

When we apply this to the left hand side of the equation, we get

$$\frac{x \frac{dy}{dx} – y}{x^2} = \sec{(x)}\tan{(x)}$$

Isolating $$\frac{dy}{dx}$$, we get

$$x \frac{dy}{dx} – y = x^2 (\sec{(x)}\tan{(x)})$$

$$x \frac{dy}{dx} = x^2 \sec{(x)}\tan{(x)} + y$$

$$\frac{dy}{dx} = \frac{x^2 \sec{(x)}\tan{(x)} + y}{x}$$

$$\frac{dy}{dx} = \frac{x^2 \sec{(x)}\tan{(x)}}{x} + \frac{y}{x}$$

We recall from the beginning that $$y = x\sec{(x)}$$. By substituting this for y in our derivative, we get

$$\frac{dy}{dx} = \frac{x^2 \sec{(x)}\tan{(x)}}{x} + \frac{x\sec{(x)}}{x}$$

Simplifying, we have

$$\frac{dy}{dx} = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

This also gives us the derivative formula for $$y = x\sec{(x)}$$

$$y’ = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

### Proof of the derivative of x sec(x) using implicit differentiation through quotient rule and the inverse secant function

Having the equation

$$y = x\sec{(x)}$$

By cross multiplying the first multiplicand of the right hand side to the left hand side of the equation, we get

$$\frac{y}{x} = \sec{(x)}$$

Equate the equation in terms of the secant function’s angle x. This will necessitate the use of the inverse secant function.

$$\sec^{-1}{\left( \frac{y}{x} \right)} = x$$

In terms of x, evaluate the implicit differentiation. On the left hand side, we’ll use the chain rule formula, which includes the quotient rule and the derivative of inverse secant, and on the right, we’ll use a basic power rule.

$$\frac{d}{dx} \left( \sec^{-1}{\left( \frac{y}{x} \right)} \right) = x$$

$$\left( \frac{1}{ \frac{y}{x} \sqrt{ \left( \frac{y}{x} \right)^2 – 1 } } \right) \cdot \left( \frac{x \frac{dy}{dx} – y}{x^2} \right) = 1$$

Isolating $$\frac{dy}{dx}$$, we get

$$\frac{ \frac{x \frac{dy}{dx} – y}{x^2} }{ \frac{y}{x} \sqrt{ \left( \frac{y}{x} \right)^2 – 1 }} = 1$$

$$\frac{x \frac{dy}{dx} – y}{x^2} = \frac{y}{x} \sqrt{ \left( \frac{y}{x} \right)^2 – 1 }$$

$$x \frac{dy}{dx} – y = x^2 \cdot \left( \frac{y}{x} \sqrt{ \left( \frac{y}{x} \right)^2 – 1 } \right)$$

$$x \frac{dy}{dx} = x^2 \left( \frac{y}{x} \sqrt{ \left( \frac{y}{x} \right)^2 – 1 } \right) + y$$

$$\frac{dy}{dx} = \frac{ \frac{x^2y}{x} \sqrt{ \left( \frac{y}{x} \right)^2 – 1 } + y }{x}$$

$$\frac{dy}{dx} = \frac{ xy \sqrt{ \left( \frac{y}{x} \right)^2 – 1 } + y }{x}$$

$$\frac{dy}{dx} = \frac{ xy \sqrt{ \left( \frac{y}{x} \right)^2 – 1 } }{x} + \frac{y}{x}$$

We recall from the beginning that $$y = x\sec{(x)}$$. By substituting this to the y of our derivative, we get

$$\frac{dy}{dx} = \frac{ x(x\sec{(x)}) \sqrt{ \left( \frac{x\sec{(x)}}{x} \right)^2 – 1 } }{x} + \frac{x\sec{(x)}}{x}$$

Simplifying, we have

$$\frac{dy}{dx} = (x\sec{(x)}) \sqrt{ \left( \frac{x\sec{(x)}}{x} \right)^2 – 1 } + \sec{(x)}$$

$$\frac{dy}{dx} = (x\sec{(x)}) \sqrt{(\sec{(x)})^2 – 1} + \sec{(x)}$$

$$\frac{dy}{dx} = (x\sec{(x)}) \sqrt{\sec^{2}{(x)} – 1} + \sec{(x)}$$

Remember that we may use a trigonometric identity here by employing $$\sec^{2}{(x)} = 1 + \tan^{2}{(x)}$$, which is known as the Pythagorean formula for tangents and secants. By utilizing this identity, we have

$$\frac{dy}{dx} = (x\sec{(x)}) \sqrt{\tan^{2}{(x)}} + \sec{(x)}$$

$$\frac{dy}{dx} = (x\sec{(x)}) \cdot (\tan{(x)}) + \sec{(x)}$$

$$\frac{dy}{dx} = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

We now get the derivative of $$y = x\sec{(x)}$$ just like in the prior two proofs.

$$y’ = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

## How to derive the x sec(x)

Before we begin the step-by-step procedure, to summarize, the derivative of $$x\sec{(x)}$$ is

### Derivative of the x sec(x) using the product rule

Step 1: Determine the two functions that will be multiplied. The first multiplicand/term shall be denoted by u, and the second by v.

$$u = x$$

$$v = \sec{(x)}$$

Step 2: Remember the derivative formula for the product rule.

$$\frac{d}{dx} f(x) = u \frac{d}{dx} v + v \frac{d}{dx} u$$

Step 3: Using the product rule formula, begin deriving $$x\sec(x)$$.

$$\frac{dy}{dx} = x \cdot \frac{d}{dx} \sec{(x)} + \sec{(x)} \cdot \frac{d}{dx} x$$

Step 4: For the derivatives of u and v, use the proper derivative formulas. In this instance, use the power rule for u‘s derivative and the derivative of secant for v‘s derivative.

$$\frac{dy}{dx} = x \cdot (\sec{(x)}\tan{(x)}) + \sec{(x)} \cdot (1)$$

Step 5: Simplify algebraically and use applicable identities or laws only when necessary.

$$\frac{dy}{dx} = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

Step 6: Finalize the answer.

$$\frac{d}{dx} x\sec{(x)} = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

## Graph of x sec(x) vs. its derivative

In the case of this function

$$f(x) = x\sec{(x)}$$

the graph is depicted as

And, as previously discussed, deriving $$f(x) = x\sec{(x)}$$ is

$$f'(x) = x\sec{(x)}\tan{(x)} + \sec{(x)}$$

which is graphically depicted as

Depicting both graphs in one, we get

Using these graphs to compare the differences between these functions, you can see that both the original function $$f(x) = x\sec{(x)}$$ and its derivative $$f'(x) = x\sec{(x)}\tan{(x)} + \sec{(x)}$$ results to the same domain of

$$\left( -\frac{5\pi}{2} , -\frac{3\pi}{2} \right) \cup \left( -\frac{3\pi}{2} , -\frac{\pi}{2} \right) \cup \left( -\frac{\pi}{2} , \frac{\pi}{2} \right) \cup \left( \frac{\pi}{2} , \frac{3\pi}{2} \right) \cup \left( \frac{3\pi}{2} , \frac{5\pi}{2} \right)$$

within the finite intervals of

$$-\frac{5\pi}{2} , \frac{5\pi}{2}$$

both of which are also within the range of

$$(-\infty,\infty)$$ or $$y | y \in \mathbb{R}$$

## Examples

The following example explains how to use the product rule formula to derive a variable multiplied by the secant of the same variable.

### EXAMPLE 1

Derive: $$f(\beta) = \beta\sec{(\beta)}$$

Solution: By applying the product rule formula, we have

Determine u and v.

$$u = \beta$$

$$v = \sec{(\beta)}$$

Remember and use the product rule formula to derive $$\beta\sec{(\beta)}$$

$$\frac{d}{d\beta} f(\beta) = u \frac{d}{d\beta} v + v \frac{d}{d\beta} u$$

$$\frac{d}{d\beta} f(\beta) = \beta \frac{d}{d\beta} \sec{(\beta)} + \sec{(\beta)} \frac{d}{d\beta} \beta$$

Derive u and v in the product rule formula

$$\frac{d}{d\beta} f(\beta) = \beta \cdot (\sec{(\beta)}\tan{(\beta)}) + \sec{(\beta)} \cdot (1)$$

Simplify algebraically and use applicable identities or laws only when necessary.

$$\frac{d}{d\beta} f(\beta) = \beta \sec{(\beta)}\tan{(\beta)} + \sec{(\beta)}$$

The final answer is:

$$\frac{d}{d\beta} \beta\tan{(\beta)} = \beta \sec{(\beta)}\tan{(\beta)} + \sec{(\beta)}$$  