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Find the derivative of the following functions. $$y=\ln \left(x^{4}+1\right)$$

Short Answer

Expert verified
Answer: The derivative of the function $$y=\ln\left(x^{4}+1\right)$$ is $$y'(x)=\frac{4x^3}{x^4+1}$$.

Step by step solution

01

Identify the functions involved

In this problem, we are given the function $$y=\ln\left(x^{4}+1\right)$$. It is a composite function involving the natural logarithm function and a simpler function inside it, $$u(x)=x^4+1$$.
02

Differentiate the inner function

We will first find the derivative of the inner function, $$u(x)$$. $$u'(x)= \frac{d}{dx}(x^4+1)=4x^3$$
03

Differentiate the outer function using the chain rule

Now we need to differentiate the composite function y = ln(u(x)). The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function. Let the outer function be $$v(u) = \ln(u)$$. The derivative of the natural logarithm function is $$\frac{1}{u}$$. Therefore, $$v'(u)=\frac{1}{u}$$
04

Apply the chain rule

Now, we can apply the chain rule to find the derivative of $$y=\ln\left(x^{4}+1\right)$$. The chain rule states that, $$y'(x)=v'(u(x))\cdot u'(x)$$ Substitute the results from Step 2 and Step 3: $$y'(x)=\frac{1}{u(x)}\cdot 4x^3=\frac{1}{x^4+1}\cdot 4x^3$$
05

Simplify the derivative

Finally, simplify the expression: $$y'(x)=\frac{4x^3}{x^4+1}$$ So, the derivative of the function $$y=\ln\left(x^{4}+1\right)$$ is $$y'(x)=\frac{4x^3}{x^4+1}$$.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain rule
The chain rule is a fundamental concept used in calculus for differentiating composite functions. Imagine you have a scenario where a function is inside another function. To find the derivative of such composite functions, the chain rule comes to our rescue with a simple formula. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

In mathematical terms, if you have a composite function \(y = f(g(x))\), the chain rule tells you that the derivative \(y'(x)\) is found by multiplying \(f'(g(x))\) with \(g'(x)\). This beautiful interplay between the inner and the outer functions allows us to break down complex derivatives into manageable chunks.

In our exercise, to differentiate \(y = \ln(x^4 + 1)\), we recognized \(x^4 + 1\) as the inner function and \(\ln(u)\) as the outer function. This identification helped us to apply the chain rule effectively by finding the derivative of each part separately and then multiplying them together.
Composite functions
Composite functions are essentially functions within functions. Think of it like stacking processes on top of each other. For example, in the function \(y = \ln(x^4 + 1)\), we have an inner function \(u(x) = x^4 + 1\) nestled within the outer function \(\ln(u)\).

Being able to identify these layers is crucial because it dictates how you approach finding the derivative, especially using the chain rule. When you encounter a composite function:
  • Identify the inner and outer functions.
  • Differentiate each function separately.
  • Use the chain rule to combine their derivatives.

This modular approach simplifies complex differentiation problems. In practice, whenever you spot a function that wraps around another—like taking the logarithm of another function—think of it in terms of composite functions.
Natural logarithm
The natural logarithm, represented by \(\ln\), is a special type of logarithm that has immense importance in calculus and beyond. It is the inverse operation of exponentiation with base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.718.

Natural logarithms have nice properties that make them convenient to work with in calculus. Notably, the derivative of \(\ln(u)\), where \(u\) is a function of \(x\), is given by \(\frac{1}{u}\). This relation is key when applying the chain rule, especially in problems involving logarithms like in our exercise.

In the case \(y = \ln(x^4 + 1)\), the derivative \(\frac{d}{du}(\ln(u)) = \frac{1}{u}\) is applied. By identifying \(x^4 + 1\) as \(u\), we smoothly work out the derivative by coupling this with the derivative of the inner function using the chain rule. Thus, mastery over natural logarithms not only aids in differentiation tasks but also enhances understanding of exponential growth and decay in various applications.

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Most popular questions from this chapter

Suppose \(f\) is differentiable on [-2,2] with \(f^{\prime}(0)=3\) and \(f^{\prime}(1)=5 .\) Let \(g(x)=f(\sin x)\) Evaluate the following expressions. a. \(g^{\prime}(0)\) b. \(g^{\prime}\left(\frac{\pi}{2}\right)\) c. \(g^{\prime}(\pi)\)

Calculating limits exactly Use the definition of the derivative to evaluate the following limits. $$\lim _{h \rightarrow 0} \frac{(3+h)^{3+h}-27}{h}$$

The population of a culture of cells after \(t\) days is approximated by the function \(P(t)=\frac{1600}{1+7 e^{-0.02 t}},\) for \(t \geq 0\) a. Graph the population function. b. What is the average growth rate during the first 10 days? c. Looking at the graph, when does the growth rate appear to be a maximum? d. Differentiate the population function to determine the growth rate function \(P^{\prime}(t)\) e. Graph the growth rate. When is it a maximum and what is the population at the time that the growth rate is a maximum?

a. Determine an equation of the tangent line and the normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. (See instructions for Exercises 73-78. b. Graph the tangent and normal lines on the given graph. \(\left(x^{2}+y^{2}\right)^{2}=\frac{25}{3}\left(x^{2}-y^{2}\right); \left(x_{0}, y_{0}\right)=(2,-1)\) (lemniscate of Bernoulli) (Graph cant copy)

Carry out the following steps. a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point. $$x^{2}+x y+y^{2}=7 ;(2,1)$$ (Graph cant copy)

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