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Magazine Chapter 3: Problem 43
Differentiate. $$y=e^{\sqrt{x-7}}$$
Short Answer
Expert verified
\(\frac{dy}{dx} = \frac{e^{\sqrt{x-7}}}{2\sqrt{x-7}}\)
Step by step solution
01
Identify the outer function and the inner function
The given function is of the form \(y = e^{u}\), where the exponent \(u\) itself is another function, in this case, \(u = \sqrt{x-7}\). Identify that \(e^{u}\) is the outer function and \sqrt{x-7} is the inner function.
02
Differentiate the outer function
Apply the chain rule. Differentiate the outer function \(e^{u}\) with respect to \(u\), which gives \(\frac{d}{du}(e^{u}) = e^{u}\).
03
Differentiate the inner function
Differentiate the inner function \(u = \sqrt{x-7}\) with respect to \(x\). First, rewrite \(u = (x - 7)^{1/2}\). Then differentiate using the power rule: \(\frac{du}{dx} = \frac{1}{2}(x - 7)^{-1/2} = \frac{1}{2\sqrt{x-7}}\).
04
Apply the chain rule
Combine the results from the previous steps. According to the chain rule, the derivative of \(y\) with respect to \(x\) is given by \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\). Using \(\frac{dy}{du} = e^{u}\) and \(\frac{du}{dx} = \frac{1}{2\sqrt{x-7}}\), we have \(\frac{dy}{dx} = e^{\sqrt{x-7}} \cdot \frac{1}{2\sqrt{x-7}}\).
05
Simplify the result
Simplify the expression to get the final answer: \(\frac{dy}{dx} = \frac{e^{\sqrt{x-7}}}{2\sqrt{x-7}}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
composite functions
A composite function is created when one function is applied to the result of another function. In mathematical terms, if we have two functions, say, \(f(x)\) and \(g(x)\), then the composite function is written as \((f \bigcirc g)(x)\) or just \(f(g(x))\). Understanding composite functions is essential when solving complex differentiation problems.
In the original exercise, \(y = e^{\textstyle \big(}\textstyle \big(\textstyle \big(x - 7\textstyle \big)\textstyle \big)^{1/2}\) can be seen as a composite function where:
In the original exercise, \(y = e^{\textstyle \big(}\textstyle \big(\textstyle \big(x - 7\textstyle \big)\textstyle \big)^{1/2}\) can be seen as a composite function where:
- The outer function, \(f(u) = e^u\)
- The inner function, \(u = (x-7)^{1/2}\)
power rule
The power rule is a basic rule in differentiation that helps find the derivative of any power of \(x\). If you have a function in the form \(y = x^n\), then the derivative with respect to \(x\) is given by \(\frac{d}{dx}(x^n) = nx^{n-1}\). This rule simplifies the differentiation process by allowing you to quickly compute the derivative of polynomial expressions.
In the context of our exercise, we use the power rule to differentiate the inner function \(u = (x-7)^{1/2}\). Rewriting it in terms of powers, we get \(u = (x-7)^{1/2}\), and differentiating this using the power rule results in:
In the context of our exercise, we use the power rule to differentiate the inner function \(u = (x-7)^{1/2}\). Rewriting it in terms of powers, we get \(u = (x-7)^{1/2}\), and differentiating this using the power rule results in:
- \frac{d}{dx}[u] = \frac{1}{2}(x-7)^{-1/2} = \frac{1}{2\textstyle \big \textstyle \big (\textstyle \big(x-7\textstyle \big)^{1/2}}
exponential functions
Exponential functions are those where the variable appears in the exponent, specifically in the form \(y = e^x\), where \(e\) is the base of the natural logarithm, approximately equal to 2.71828. These functions have the unique property that their rate of change (derivative) is proportional to the function itself.
In our given exercise, you encountered an exponential function with a composite argument, \(y = e^{u}\). When differentiating \(e^u\) with respect to \(u\), we get \( \frac{d}{du}(e^u) = e^u \). This property is used in the step where we differentiated \(e^u\) to be \( \frac{d}{du}(e^{\textstyle \big(\textstyle \big(\textstyle \big(x-7\textstyle \big)^{1/2}}\).
In our given exercise, you encountered an exponential function with a composite argument, \(y = e^{u}\). When differentiating \(e^u\) with respect to \(u\), we get \( \frac{d}{du}(e^u) = e^u \). This property is used in the step where we differentiated \(e^u\) to be \( \frac{d}{du}(e^{\textstyle \big(\textstyle \big(\textstyle \big(x-7\textstyle \big)^{1/2}}\).
derivatives
Derivatives represent the rate at which a function changes as its input changes. Symbolically, the derivative of \(y\) with respect to \(x\) is written as \( \frac{dy}{dx} \). It is a fundamental concept in calculus and is used to determine slopes of curves, velocities, accelerations, and more.
In this exercise, we use the chain rule alongside the basic rules of differentiation to find the derivative of a composite exponential function. This process involves differentiating the outer function first, then multiplying by the derivative of the inner function:
In this exercise, we use the chain rule alongside the basic rules of differentiation to find the derivative of a composite exponential function. This process involves differentiating the outer function first, then multiplying by the derivative of the inner function:
- From Step 3, we have \( \frac{du}{dx} = \frac{1}{2 \textstyle \big{\textstyle \big(\textstyle \big(x-7\textstyle \big)}^{1/2}}
- From step 4, the outer function differentiated resulted in \) e^{u} \(
- Multiply them together and simplify:
\) \frac{dy}{dx} = e^{ \textstyle \big(\textstyle \big{\textstyle \big(x-7\textstyle \big)}^{1/2}\( \frac{ 1}{ 2 }{\textstyle \big(\textstyle \big({ x-7\textstyle \big } ^{1/2}\) To get the final simplified derivative \( \frac{dy}{dx} = \frac{e^\textstyle \big({ \textstyle \big (x-7 }^{ 1/2 }}{ 2\textstyle \big({ \textstyle \big(x-7)}^{1/2}}\).
