Question

Find the derivative of the following functions. $$y=x \sin x$$

Short answer

Answer: The derivative of the given function is $$\frac{dy}{dx} = -2x\ln x^2$$.

Step-by-step solution

Find the derivative of $$x^2$$

To find the derivative of $$x^2$$, we use the power rule of differentiation. The power rule states that if $$y = x^n$$, then $$y' = nx^{n-1}$$. In our case, $$n=2$$, so the derivative is: $$\frac{d}{dx}(x^2) = 2x^{2-1} = 2x$$

Find the derivative of $$(1-\ln x^2)$$

To find the derivative of $$(1-\ln x^2)$$, we need to first find the derivative of $$\ln x^2$$ and then apply the subtraction rule of differentiation. The derivative of $$\ln x^2$$ can be found using the chain rule, which states that if $$y = \ln (u(x))$$, then $$y' = \frac{1}{u(x)} \cdot u'(x)$$. In our case, we have: $$u(x) = x^2 \Rightarrow u'(x) = 2x$$ $$\frac{d}{dx}(\ln x^2) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}$$ Now we can find the derivative of $$(1-\ln x^2)$$ using the subtraction rule: $$\frac{d}{dx}(1-\ln x^2) = \frac{d}{dx}(1) - \frac{d}{dx}(\ln x^2) = 0 - \frac{2}{x} = -\frac{2}{x}$$

Apply the product rule to find the derivative of the given function

Now, we will find the derivative of the given function $$y=x^2\left(1-\ln x^{2}\right)$$ by applying the product rule using the derivatives we found in Steps 1 and 2: $$\frac{dy}{dx} = (2x)\left(1-\ln x^2\right) + \left(x^2\right)\left(-\frac{2}{x}\right)$$ Simplified, this gives us: $$\frac{dy}{dx} = 2x(1-\ln x^2) - 2x = 2x(1-\ln x^2 - 1) = -2x\ln x^2$$ So, the derivative of the given function is: $$\frac{dy}{dx} = -2x\ln x^2$$

Key concepts

Power Rule

The power rule is a fundamental concept in calculus that makes finding derivatives a breeze. It's incredibly useful when dealing with polynomial functions. The wonderful thing about the power rule is its simplicity: To differentiate a function of the form \( y = x^n \), you merely bring the exponent down as a coefficient and decrease the exponent by one. This means that the derivative, \( y' \), is calculated as \( y' = nx^{n-1} \).

• Example: If you have \( y = x^2 \), applying the power rule gives you \( y' = 2x^{2-1} = 2x \).
• This basic rule is applicable to any real number \( n \), whether positive, negative, or fractional.
• The power rule is not just for handling simple terms individually but also becomes quite handy in breaking down more complex expressions where power functions are involved.

The power rule is a fundamental building block for other rules in calculus, helping to represent growth rates and changes in algebraic expressions swiftly. Understanding the power rule really sets the foundation for tackling more complicated forms of differentiation in calculus.

Chain Rule

Another key element in calculus is the chain rule, which allows us to differentiate composite functions. Composite functions occur when you have one function nested inside another. The chain rule comes into play beautifully here, allowing us to unravel these layers of functions.Consider a function \( y = f(g(x)) \), where \( g(x) \) is inside \( f(x) \). The chain rule tells us that to find the derivative, we differentiate the outer function and multiply it by the derivative of the inner function, expressed as \( y' = f'(g(x)) \cdot g'(x) \).

• Example: For \( y = \ln(x^2) \), identify the inner function \( u(x) = x^2 \) and find \( u'(x) = 2x \). The outer function is \( \ln(u) \), giving us the derivative \( \frac{d}{du}(\ln u) = \frac{1}{u} \). Combining these gives \( y' = \frac{1}{x^2} \cdot 2x = \frac{2}{x} \).
• The chain rule is powerful for functions where one variable affects another by feeding into it, common in real-world scenarios like growth rates and physical models.
• It enables breaking down complex functions into manageable parts, interacting cleverly with other rules like the power rule.

Mastering the chain rule is essential for working with complex and deeply nested functions, providing the skills needed to dissect and handle intricate expressions in calculus.

Product Rule

When dealing with products of two functions in differentiation, the product rule is your go-to method. It's designed to find the derivative of expressions that have two differentiable functions being multiplied together. The product rule formula is: If \( y = u(x) \cdot v(x) \), then \( y' = u'(x) \cdot v(x) + u(x) \cdot v'(x) \).

• Example: For the function \( y = x^2(1 - \ln x^2) \), you identify \( u(x) = x^2 \) and \( v(x) = (1 - \ln x^2) \).
• The derivatives are \( u'(x) = 2x \) and \( v'(x) = -\frac{2}{x} \).
• Applying the product rule gives \( y' = (2x)(1-\ln x^2) + (x^2)(-\frac{2}{x}) \), which simplifies to the expression \( -2x\ln x^2 \).

The product rule is indispensable when functions multiply together, helping you maintain the integrity of both components. It's especially useful in physics and engineering, where multipliers often represent independent factors influencing a system. Understanding and applying the product rule aids in comprehending how changes in one part of an equation affects the whole, paving the way for higher-level problem solving in calculus.