Question

In Exercises \(1-8,\) use the given substitution and the Chain Rule to find \(d y / d x\) $$y=\tan \left(2 x-x^{3}\right), u=2 x-x^{3}$$

Short answer

The derivative of \(y=\tan\(2x-x^3\) is \(dy/dx=\sec^2(2x-x^3) \cdot (2-3x^2)\).

Step-by-step solution

Differentiate \(y=\tan(u)\) with respect to 'u' using the chain rule

The chain rule of derivatives needs to be applied here. The derivative of \(y=\tan(u)\) is \(dy/du = \sec^2(u)\).

Differentiate \(u=2x-x^3\) with respect to 'x'

By applying the power rule for derivatives, \(u'(x) = 2-3x^2\). Here, the power rule is applied separately to each term of the function \(u=x\).

Substitute \(du/dx\) into the chain rule

Combine both obtained expressions using the chain rule. The chain rule is expressed as \(dy/dx = dy/du × du/dx\). So substituting given values gives us \(dy/dx=\sec^2(u) \cdot (2-3x^2)\). Plugging the value of \(u\) back into the equation we get \(dy/dx=\sec^2(2x-x^3) \cdot (2-3x^2)\).

Key concepts

Derivative of tan(u)

Understanding the derivative of the tangent function with respect to a variable is a fundamental concept in calculus. The tangent function, denoted by \(tan(u)\), is one of the six trigonometric functions. To find its derivative, we use the fact that the tangent can be expressed in terms of sine and cosine functions: \(tan(u) = \frac{sin(u)}{cos(u)}\). However, a more direct approach uses trigonometric identities and defines that the derivative of \(tan(u)\) with respect to \(u\) is \(sec^2(u)\).
In practical terms, whenever you have a function that involves the tangent of an expression, like in our exercise \(y=tan(2x-x^3)\), you will apply this knowledge. By differentiating \(tan(u)\) with respect to \(u\), you obtain \(\frac{dy}{du} = sec^2(u)\). This expression becomes a part of the chain rule when you're dealing with composite functions, which occurs frequently in calculus problems involving trigonometric functions.

Power rule for derivatives

The power rule is an essential tool for differentiating functions where the variable has an exponent. In essence, the power rule states that for any function of the form \(f(x) = x^n\), where \(n\) is a real number, the derivative of the function with respect to \(x\) is \(f'(x) = nx^{n-1}\).

Applying the Power Rule

When dealing with a polynomial like \(u = 2x - x^3\) in our exercise, you apply the power rule term-by-term. For the linear term \(2x\), the power is implicitly one, and following the rule, we get a derivative of \(2\). For the term \(-x^3\), we apply the power rule and multiply the exponent by the coefficient, then decrease the exponent by one, to find the derivative as \(-3x^2\). Adding these results together, we find the derivative of \(u\) with respect to \(x\) as \(u'(x) = 2 - 3x^2\), which is crucial for continuing the process of taking the derivative of the composite function using the chain rule.

Differentiating composite functions

Composite functions arise when you have a function inside another function. Differentiation of such functions requires what is known as the chain rule. The chain rule is a way to differentiate composite functions, and it can be thought of as finding the derivative of the outer function multiplied by the derivative of the inner function.

Implementing the Chain Rule

In the given exercise, \(y = \tan(2x - x^3)\), the inner function is \(u = 2x - x^3\) and the outer function is the tangent function of \(u\). According to the chain rule, to find \(\frac{dy}{dx}\), you multiply the derivative of the outer function with respect to the inner function \(\frac{dy}{du}\) by the derivative of the inner function with respect to \(x\) \(\frac{du}{dx}\).
After determining \(\frac{dy}{du} = \sec^2(u)\) and \(\frac{du}{dx} = 2 - 3x^2\), you combine both to get the derivative of \(y\) with respect to \(x\) as \(\frac{dy}{dx} = \sec^2(2x - x^3) \cdot (2 - 3x^2)\). This method simplifies the process of differentiation even when dealing with more complex expressions, enabling us to systematically tackle calculus problems involving derivatives.