Question

Evaluate the following derivatives. $$\frac{d}{dx}(x\ln x^3)$$

Short answer

Question: Find the derivative of the given function with respect to x: $$y=x\ln x^3$$

Step-by-step solution

Identify the functions in the product rule

The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. In our case, we have: $$f(x)=x$$ and $$g(x)=\ln x^3$$

Differentiate f(x)

To differentiate f(x) = x with respect to x, we will simply apply the power rule, which states that the derivative of x^n is nx^(n-1). In our case, n = 1, so the derivative is 1x^(1-1) = 1x^0 =1. So we have : $$f^{\prime }(x)=1$$

Differentiate g(x) using the chain rule

To find the derivative of g(x) = ln x^3, we will use the chain rule because the logarithm's argument is x^3. The chain rule states that the derivative of a composite function h(g(x)) is the derivative of h with respect to g(x) times the derivative of g(x) with respect to x. In our case, h(u) = ln(u) and g(x) = x^3. So we have: $$g^{\prime }(x)=\frac{d}{du}(\ln u)\cdot \frac{d}{dx}(x^3)$$ To differentiate h(u) = ln(u) with respect to u, the derivative is 1/u. Since u = g(x) = x^3, we have: $$\frac{d}{du}(\ln u)=\frac{1}{x^3}$$ Next, we differentiate g(x) = x^3 with respect to x using the power rule: $$\frac{d}{dx}(x^3)=3x^2$$ Now, we can find the derivative g'(x) using the chain rule: $$g^{\prime }(x)=\frac{1}{x^3}\cdot 3x^2$$ This simplifies to: $$g^{\prime }(x)=\frac{3}{x}$$

Apply the product rule to find the derivative

Now that we have the derivatives of f(x) and g(x), we can apply the product rule to find the derivative of the given function: $$\frac{d}{dx}(x\ln x^3)=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$$ Substitute the expressions for f'(x), g(x), f(x), and g'(x): $$\frac{d}{dx}(x\ln x^3)=(1)(\ln x^3)+(x)\left(\frac{3}{x}\right)$$ Now simplify: $$\frac{d}{dx}(x\ln x^3)=\ln x^3+3$$ Therefore, the derivative of the given function with respect to x is: $$\frac{d}{dx}(x\ln x^3)=\ln x^3+3$$

Key concepts

Derivative

In calculus, a derivative represents the rate at which a function changes as its input changes. Consider it as the slope of the tangent line at any point on the graph of the function. For example, given a function $f(x)$, its derivative $f^{\prime }(x)$ provides the rate of change of $f$ with respect to $x$.

Understanding derivatives is crucial because they are foundational to many concepts, such as finding maxima and minima of functions, analyzing graphs, and solving real-world problems where change is involved. When dealing with derivatives, always look for rules that simplify calculations like the product, chain, and power rules. These rules allow you to work with more complex functions efficiently.

In our example, we focused on differentiating $x\ln x^3$ using the product rule, combining it with other rules to simplify the derivative steps and obtain $\ln x^3+3$.

Chain Rule

The chain rule is a powerful tool in differentiation, especially for composite functions. When a function is nested inside another, the chain rule helps us find the derivative conveniently. It states:

• If you have a composite function $h(x)=f(g(x))$, then the derivative $h^{\prime }(x)$ is $f^{\prime }(g(x))\cdot g^{\prime }(x)$.

This rule requires you to take the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. This method simplifies handling expressions like $\ln (x^3)$ in calculus.

In our case, the chain rule was used to differentiate $\ln (x^3)$. Here:

• The outer function $f(u)=\ln u$ has a derivative $\frac{1}{u}$.
• The inner function $g(x)=x^3$, differentiated, gives $3x^2$.

Applying the chain rule: $g^{\prime }(x)=\frac{1}{x^3}\cdot 3x^2=\frac{3}{x}$.

This step ensures we accurately capture how changes in $x$ affect the entire composite function, leading to more precise calculus operations.

Power Rule

The power rule provides a straightforward method to find the derivative of any polynomial function. For any term $x^n$, the power rule dictates that its derivative is $n{x}^{n-1}$.

This rule applies simply and efficiently:

• If $f(x)=x^n$, then $f^{\prime }(x)=n{x}^{n-1}$.

You reduce the exponent by 1 and multiply by the original exponent, making it a go-to method for polynomials.

In our example, the power rule was essential in differentiating $f(x)=x$ and $x^3$. Here:

• For $x$, $n=1$, leading to $1\times x^0=1$.
• For $x^3$, $n=3$, giving $3x^2$.

This simplicity and utility make it a fundamental tool in finding derivatives quickly without delving into more complex rules right away. It forms the backbone of early calculus work and solves common differentiation problems efficiently.