Question

Compute the following derivatives using the method of your choice. $$\frac{d}{d x}\left(\cos \left(x^{2 \sin x}\right)\right)$$

Short answer

Question: Find the derivative of the function $$y = \cos(x^{2\sin x})$$ with respect to x. Answer: The derivative of the given function, $$\frac{dy}{dx}$$, is $$-\sin(x^{2\sin x})\cdot4\sin x \cdot x^{2\sin x - 1}\cdot \cos x$$.

Step-by-step solution

Derivative of the Outer Function

As previously stated, the outer function is $$f(u)=\cos u$$. We now need to find $$\frac{d}{d u}(f(u))$$: $$ \frac{d}{du}\left(\cos u\right) = -\sin u $$

Derivative of the Inner Function

Now, we need to find the derivative of $$g(x)=x^{2\sin x}$$ with respect to x. To do this, we will use the power rule combined with the chain rule: $$ \frac{d}{dx}\left(x^{2 \sin x}\right) = 2\sin x \cdot x^{2\sin x - 1}\cdot\frac{d}{dx}\left(2\sin x\right) $$ The derivative of $$2\sin x$$ with respect to x is $$2\cos x$$, so we can substitute this into our expression from above: $$ \frac{d}{dx}\left(x^{2 \sin x}\right) = 2\sin x \cdot x^{2\sin x - 1}\cdot(2\cos x) $$

Apply the Chain Rule

Now that we have the derivatives of the outer and inner functions, we can apply the chain rule to find the derivative of the given function: $$ \frac{d}{dx}\left(\cos\left(x^{2\sin x}\right)\right)= -\sin\left(x^{2\sin x}\right)\cdot2\sin x \cdot x^{2\sin x - 1}\cdot(2\cos x) $$

Final Answer

The derivative of the given function, $$\frac{d}{d x}\left(\cos \left(x^{2 \sin x}\right)\right)$$, is: $$ -\sin\left(x^{2\sin x}\right)\cdot4\sin x \cdot x^{2\sin x - 1}\cdot \cos x $$

Key concepts

Chain Rule

The Chain Rule is a fundamental concept in calculus for finding the derivative of composite functions. When dealing with a composite function, such as \( f(g(x)) \), the chain rule helps us break things down.To apply the chain rule, you need to:

• Identify the outer function and the inner function. In our exercise, the outer function is \( \,\cos(u)\, \) and the inner function is \( \,x^{2\sin x}\, \).
• Find the derivative of the outer function with respect to the inner function, which in this case yields \( -\sin(u) \).
• Next, compute the derivative of the inner function \( x^{2\sin x} \) with respect to \( x \), which involves both the power rule and chain rule, details explained later.

Finally, multiply these derivatives to get the final derivative of the composite function. The chain rule's beauty lies in its ability to simplify the differentiation of functions nested within each other, just as we saw in our example.

Power Rule

The Power Rule is crucial for differentiating functions involving powers of \( x \). It states that if you have a function in the form \( x^n \), its derivative is \( n \times x^{n-1} \).In the exercise, the inner function is \( x^{2\sin x} \). Here's where it gets interesting:

• The exponent itself, \( 2\sin x \), is also a function of \( x \), requiring us to combine the power rule with the chain rule for proper differentiation.
• The initial derivative expression is \( 2\sin x \times x^{2\sin x - 1} \).
• Because \( 2\sin x \) changes as \( x \) changes, its derivative also affects the calculation, leading us to \( 2\cos x \).

By applying these steps, we accurately determine a complex derivative involving variable exponents. The power rule, especially when used with functions as exponents, showcases its adaptability and importance in calculus.

Trigonometric Functions

Trigonometric functions, like sine and cosine, are instrumental in calculus, particularly in differentiation. They are commonly seen in problems involving periodic phenomena or geometric relations.For differentiation purposes, memorize the basic derivatives:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).

In our exercise, trigonometric functions appear both in the inner and outer components:

• The outer function is \( \cos(u) \), whose derivative is \( -\sin(u) \).
• The inner function involves \( 2\sin(x) \) as an exponent, which alters the base's differentiation.

Understanding and calculating these derivatives accurately is crucial when dealing with composite functions containing trigonometric parts, as they significantly influence the shape and interpretation of a function.