Question

Use the Chain Rule combined with other differentiation rules to find the derivative of the following functions. $$y=e^{x^{2}+1} \sin x^{3}$$

Short answer

Question: Determine the derivative of the function \(y = e^{x^2 + 1}\sin x^3\) with respect to \(x\). Answer: The derivative, \(y'\), is: $$y' = 2xe^{x^2+1}\sin x^3 + 3x^2e^{x^2+1}\cos x^3$$

Step-by-step solution

Differentiate \(e^{x^2+1}\) using Chain Rule

To differentiate \(e^{x^2+1}\), first, let's call \(u = x^2 + 1\). So our function becomes \(e^u\). Now we'll find the derivative of \(e^u\) with respect to \(x\) using chain rule: \((e^u)' = e^u \cdot u'\). Now we need to find the derivative of \(u\) with respect to \(x\), which is \(u' = (x^2 + 1)' = 2x\). So the derivative of \(e^{x^2+1}\) is: $$\frac{d}{dx} e^{x^2+1} = e^{x^2+1} \cdot 2x$$

Differentiate \(\sin x^3\) using Chain Rule

Similar to step 1, let's say \(v = x^3\). So our function becomes \(\sin v\). Now, we'll find the derivative of \(\sin v\) with respect to \(x\) using chain rule: \((\sin v)' = \cos v \cdot v'\). Now we need to find the derivative of \(v\) with respect to \(x\), which is \(v' = (x^3)' = 3x^2\). So the derivative of \(\sin x^3\) is: $$\frac{d}{dx} \sin x^3 = \cos x^3 \cdot 3x^2$$

Apply Product Rule to find derivative of y

Now that we have the derivatives of both functions, we will apply the product rule to find the derivative of \(y = e^{x^2 + 1}\sin x^3\). Recall that the product rule states that \((f(x)g(x))' = f'(x)g(x) + f(x)g'(x)\). Here, \(f(x) = e^{x^2 + 1}\) and \(g(x) = \sin x^3\). Our derivatives for \(f'(x)\) and \(g'(x)\) are given: $$f'(x) = e^{x^2+1} \cdot 2x$$ $$g'(x) = \cos x^3 \cdot 3x^2$$ Now applying the product rule, we have: $$y' = f'(x)g(x) + f(x)g'(x) = (e^{x^2+1} \cdot 2x)(\sin x^3) + (e^{x^2+1})(\cos x^3 \cdot 3x^2)$$ Finally, we can simplify our result: $$ y' = 2xe^{x^2+1}\sin x^3 + 3x^2e^{x^2+1}\cos x^3 $$

Key concepts

Product Rule Differentiation

The Product Rule is an essential differentiation technique used when dealing with the product of two functions, which is frequently encountered in calculus. To understand the Product Rule, let's denote two differentiable functions as \(f(x)\) and \(g(x)\). The Product Rule states that the derivative of their product is given by \( (f(x)g(x))' = f'(x)g(x) + f(x)g'(x) \).

When applying this rule, the first step is to differentiate both functions independently. Afterwards, the products of the original function and the derivative of the other function are summed to give the final result. It’s like saying, “differentiate the first function and multiply by the second, then add the first function multiplied by the derivative of the second.”

In practice, it's important to keep the original functions intact while taking the derivatives, and not to simplify prematurely as it can lead to mistakes or a more difficult differentiation process. Furthermore, if either one of the functions has its own composite structure, such as being an exponential or trigonometric function, we must first apply their respective differentiation rules, such as the Chain Rule, before applying the Product Rule.

Exponential Function Derivation

Differentiation of exponential functions with base \(e\), the natural logarithm base, is very straightforward. The derivative of \(e^x\) with respect to \(x\) is simply \(e^x\). However, when dealing with an exponential function that includes a composite function, like \(e^{u(x)}\), we must apply the Chain Rule.

The Chain Rule helps us differentiate composite functions and is stated as \(\frac{d}{dx} [f(g(x))] = f'(g(x)) \cdot g'(x)\). When \(f(x) = e^x\) and \(g(x)\) is any other function of \(x\), the derivative is \((e^{g(x)})' = e^{g(x)} \cdot g'(x)\). This formulation is particularly useful when handling more complex expressions where the exponent is not merely \(x\), but a function of \(x\).

One critical piece of advice for students is to clearly denote the inner function, often labeled as \(u\), and to calculate its derivative separately before applying it to the differentiation of the whole function. Remember to apply the Chain Rule correctly to avoid common mistakes, such as forgetting to multiply by the derivative of the inner function.

Trigonometric Function Derivation

Trigonometric functions such as sine, cosine, and tangent, among others, have their specific derivatives that must be memorized. The derivative of \(\sin x\) is \(\cos x\), and the derivative of \(\cos x\) is \(\-\sin x\). These derivatives are the basis when working with trigonometric functions in calculus.

Similar to exponential functions, when a trigonometric function involves a composite function, the Chain Rule must be applied. For example, the derivative of \(\sin(g(x))\) is \(\cos(g(x)) \cdot g'(x)\), applying the derivative of the outer function \(\sin x\) to the inner function \(g(x)\) and then multiplying by the derivative of \(g(x)\).

To successfully differentiate trigonometric functions, especially when they occur as part of a more complex function, students are advised to first identify and differentiate the inner function before applying the trigonometric derivative. This helps prevent errors and ensures that the Chain Rule is correctly used in conjunction with trigonometric derivatives. Careful and methodical application of these rules is key to solving differentiation problems involving trigonometric functions.