Question

Differentiate the following functions. $$ u=\tan ^{2} a x $$

Short answer

Question: Find the derivative of the function $$u = \tan^2(ax)$$ with respect to $$x$$. Answer: The derivative of the function $$u = \tan^2(ax)$$ with respect to $$x$$ is $$\frac{du}{dx} = 2\tan(ax) \cdot a\sec^2(ax)$$.

Step-by-step solution

Identify the chain rule formula.

The chain rule states that for a composite function $$y = f(g(x))$$, the derivative with respect to x is: $$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}, $$ where $$u = g(x)$$ and $$y = f(u)$$. In this case, $$u = \tan^2(ax)$$, we can identify that the function has the form $$u=\left(\tan(ax)\right)^2$$, so we will apply the chain rule by taking the derivative of the inner function (the argument of the tan) and the outer function (the square operation).

Differentiate the outer function.

Now we need to differentiate the outer function, that is, $$u(v)=v^2$$ with respect to the variable $$v$$: $$ \frac{du}{dv} = 2v, $$ where $$u=\tan^2(ax)$$ and $$v=\tan(ax)$$.

Differentiate the inner function.

We need to differentiate the inner function, that is, $$v(x)=\tan(ax)$$ with respect to the variable $$x$$: $$ \frac{dv}{dx} = a\sec^2(ax), $$ where $$\sec (\theta)$$ is the reciprocal of the cosine function, so $$\sec(\theta)=\frac{1}{\cos(\theta)}$$.

Apply the chain rule.

Now we have both parts necessary for applying the chain rule. Substitute the derivative of the outer and inner functions in the chain rule formula: $$ \frac{du}{dx} = \frac{du}{dv} \cdot \frac{dv}{dx} = 2v \cdot a\sec^2(ax), $$ recall that $$v=\tan(ax)$$, so the final answer for the derivative of $$u$$ with respect to $$x$$ is $$ \frac{du}{dx} = 2\tan(ax) \cdot a\sec^2(ax). $$

Key concepts

Chain Rule

The chain rule is an essential tool in calculus for finding the derivative of composite functions. It allows us to differentiate more complex functions by breaking them down into simpler parts. Imagine you have a function that is composed of two or more functions; the chain rule helps us by instructing us to differentiate each part step by step.

Let's say we have a function defined as \(y = f(g(x))\). Here, the function \(g(x)\) is nested inside the function \(f(x)\). To find the derivative of \(y\) with respect to \(x\), the chain rule tells us to first find the derivative of \(f\) with respect to \(g\), denoted as \(\frac{dy}{du}\), and then multiply it by the derivative of \(g\) with respect to \(x\), written as \(\frac{du}{dx}\). Together, they give us \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).

This might sound complicated, but the chain rule simply breaks down the differentiation process into manageable steps, making it easier to tackle composite functions.

Composite Functions

A composite function is a function made up by combining two or more functions. It's like a mathematical nesting doll, where one function is "inside" another.

When dealing with composite functions, it's crucial to identify which part of the function is the "inner function" and which part is the "outer function." For the problem \(u = \tan^2(ax)\), the inner function is \(\tan(ax)\), and the outer function is the squaring operation \((v^2)\).

Understanding composite functions is vital because they frequently appear in calculus problems. Once you can identify the inner and outer layers, applying rules like the chain rule becomes much more manageable.

Trigonometric Functions

Trigonometric functions are fundamental in calculus and play a key role in the study of triangles and periodic phenomena. They include sine, cosine, tangent, and their reciprocals - cosecant, secant, and cotangent. In our exercise, the tangent function \(\tan(x)\) and its reciprocal \(\sec(x)\) pop up.

The tangent function, \(\tan(x)\), is defined as the ratio of the sine of an angle to the cosine of the same angle, \(\tan(x) = \frac{\sin(x)}{\cos(x)}\). Its derivative, \(\frac{d}{dx}[\tan(x)] = \sec^2(x)\), is crucial because it appears in many calculus problems.

The secant function, defined as \(\sec(x) = \frac{1}{\cos(x)}\), is important when taking derivatives of tangent as it simplifies the expressions. Understanding these trigonometric functions and how they interrelate helps greatly when differentiating functions involving trigonometry.

• Remember, trigonometric functions often cycle in and out of calculus problems, so getting comfortable with their derivatives is key.