Chapter 2: Problem 12
Use \(f^{\prime}(x)=\lim _{h \rightarrow 0}[f(x+h)-f(x)] / h\) to find the derivative at \(x\). \(g(x)=x^{4}+x^{2}\)
Short Answer
Expert verified
The derivative of \( g(x) \) is \( 4x^3 + 2x \).
Step by step solution
01
Write the Difference Quotient
The difference quotient for the function can be written based on the provided formula for the derivative: \[ f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{g(x+h) - g(x)}{h} \] For the given function \( g(x) = x^4 + x^2 \), we need to find \( g(x+h) \).
02
Compute \( g(x+h) \)
Substitute \( x+h \) into \( g(x) \): \[ g(x+h) = (x+h)^4 + (x+h)^2 \] Use the binomial expansion to find \((x+h)^4\) and \((x+h)^2\).
03
Expand \((x+h)^4\) and \((x+h)^2\)
Using the binomial formula:\[(x+h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4\]\[(x+h)^2 = x^2 + 2xh + h^2\]
04
Simplify \( g(x+h) - g(x) \)
Now calculate \( g(x+h) - g(x) \): \[ g(x+h) - g(x) = (x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4) + (x^2 + 2xh + h^2) - (x^4 + x^2) \]Bring like terms together and simplify to obtain:\[ 4x^3h + 6x^2h^2 + 4xh^3 + h^4 + 2xh + h^2 \]
05
Factor out \( h \)
We factor \( h \) from each term in the expression:\[ h(4x^3 + 6x^2h + 4xh^2 + h^3 + 2x + h) \]
06
Compute the Limit as \( h \to 0 \)
Apply the limit as \( h \) approaches 0:\[ \lim_{h \to 0} h(4x^3 + 6x^2h + 4xh^2 + h^3 + 2x + h) = 4x^3 + 2x \]This is because all terms involving \( h \) go to zero.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Difference Quotient
The difference quotient is a central concept in calculus, used to find the instantaneous rate of change of a function, or its derivative. The formula for the difference quotient is given as:\[f^{\prime}(x) = \lim _{h \rightarrow 0} \frac{f(x+h) - f(x)}{h}\]Here's how it works:
- You take the function value at \(x\) and at \(x + h\).
- Subtract these function values to see how much the function changes over the interval \([x, x+h]\).
- Divide by \(h\) to find the average rate of change, or slope, of the function over that interval.
- Finally, take the limit as \(h\) approaches zero.This gives you the actual instantaneous rate of change, or the derivative.
Binomial Expansion
Binomial expansion is a technique used to expand expressions that are raised to a power. It's based on the binomial theorem, which allows us to express powers of a sum in terms of sums and products of the components. In our case, the binomial expansion is used to simplify expressions like \[(x + h)^4\] and \[(x + h)^2\]. Using the binomial theorem:
- \((x + h)^4 = x^4 + 4x^3h + 6x^2h^2 + 4xh^3 + h^4 \)
- \((x + h)^2 = x^2 + 2xh + h^2\)
Limit of a Function
The limit of a function is a fundamental concept in calculus, used to define the behavior of a function as its input approaches a certain value. Limits are crucial for defining derivatives and integrals.In our exercise, the limit:\[\lim_{h \to 0} h(4x^3 + 6x^2h + 4xh^2 + h^3 + 2x + h) = 4x^3 + 2x\]shows us what happens to the result of the difference quotient as \(h\) becomes incredibly small, approaching zero.Important aspects of limits include:
- The ability to understand the behavior of functions at points that might not be explicitly defined.
- Helps deduce the derivative, highlighting how small changes in input affect the function's output.