Problem 33
Question
Find the derivatives of the function. $$y=x^{9 / 4}+e^{-2 x}$$
Step-by-Step Solution
Verified Answer
\(\frac{dy}{dx} = \frac{9}{4}x^{5/4} - 2e^{-2x}\)
1Step 1: Differentiate the First Term
The first term of the function is \(x^{9/4}\). Use the power rule for differentiation, which states that if \(y = x^n\), then \(\frac{dy}{dx} = nx^{n-1}\). For \(x^{9/4}\), differentiate it to get \(\frac{9}{4}x^{9/4 - 1} = \frac{9}{4}x^{5/4}\).
2Step 2: Differentiate the Second Term
The second term of the function is \(e^{-2x}\). Use the chain rule for differentiation. The derivative of \(e^{f(x)}\) is \(f'(x)e^{f(x)}\). Here, \(f(x)=-2x\), and \(f'(x)=-2\). The derivative is \(-2e^{-2x}\).
3Step 3: Combine the Derivatives
Now combine the derivatives of both terms. The derivative of the entire function \(y = x^{9/4} + e^{-2x}\) is the sum of the derivatives found in the previous steps. So, the derivative \(\frac{dy}{dx} = \frac{9}{4}x^{5/4} - 2e^{-2x}\).
Key Concepts
Power RuleChain RuleDifferentiationExponential Function
Power Rule
The power rule is a fundamental tool in differentiation used to find the derivative of functions of the form \(x^n\). This rule simplifies the process by guiding us to "bring down" the exponent as a coefficient and then reduce the original exponent by one.
- If you have a function \(y = x^n\), its derivative is \(\frac{dy}{dx} = nx^{n-1}\).
- For example, differentiating \(x^{9/4}\) using the power rule results in \(\frac{9}{4}x^{9/4 - 1} = \frac{9}{4}x^{5/4}\).
Chain Rule
The chain rule is an essential method in calculus for differentiating composite functions. It allows you to break down complicated expressions and differentiate them step by step.
- Consider a function \(y = g(f(x))\), where you must first differentiate the outer function, \(g\), with respect to the inner function, \(f\).
- The chain rule formula is \( (g(f(x)))' = g'(f(x))\cdot f'(x) \).
- In our example, \(e^{-2x}\), the challenges are tackled by first identifying \(f(x) = -2x\) and \(g(u) = e^u\), resulting in a derivative of \(-2e^{-2x}\).
Differentiation
Differentiation is the cornerstone of calculus, aimed at finding the rate at which a function changes at any given point. This process is fundamental in understanding and analyzing mathematical models of real-world situations.
- At its core, differentiation involves applying rules, like the power rule and chain rule, to compute derivatives.
- The derivative of a function, represented as \(\frac{dy}{dx}\), provides insight into the function's instantaneous rate of change.
- For the function \(y = x^{9/4} + e^{-2x}\), we differentiate each term separately and combine the results, obtaining \(\frac{9}{4}x^{5/4} - 2e^{-2x}\).
Exponential Function
Exponential functions are unique in calculus because they involve constants raised to a variable power, often appearing in the form \(e^{f(x)}\). These functions have distinct rules for differentiation that require careful application of techniques like the chain rule.
- The exponential function \(e^x\) has a unique property where its derivative is itself, \(e^x\).
- However, for \(e^{f(x)}\), we need to apply the chain rule: differentiate \(f(x)\) first and then multiply by the exponential term.
- In the problem example, \(e^{-2x}\) demands this method, as we must also differentiate the inner function \(-2x\), resulting in \(-2e^{-2x}\).
Other exercises in this chapter
Problem 33
Find \(y^{\prime \prime}\) if a. \(y=\csc x\) b. \(y=\sec x\)
View solution Problem 33
Show that the line \(y=m x+b\) is its own tangent line at any point \(\left(x_{0}, m x_{0}+b\right)\).
View solution Problem 34
Find the derivative of \(y\) with respect to the appropriate variable. $$y=\tan ^{-1}(\ln x)$$
View solution Problem 34
Find \(d y\). $$y=\ln \left(\frac{x+1}{\sqrt{x-1}}\right)$$
View solution