Tutorial Exercise Find the derivative of the function. h(x) = ex³-x+8 Step 1 For the function h(x) = ex³-x+8, note that the derivative of h. is a composite exponential function where the exponent is an expression. Hence, the generalized rule dreu If we think of ex-x+8 as e", then u is a differentiable function of x and u = Apply the generalized rule to find the derivative of e". h'(x) = [e"] du du will be used to find
Tutorial Exercise Find the derivative of the function. h(x) = ex³-x+8 Step 1 For the function h(x) = ex³-x+8, note that the derivative of h. is a composite exponential function where the exponent is an expression. Hence, the generalized rule dreu If we think of ex-x+8 as e", then u is a differentiable function of x and u = Apply the generalized rule to find the derivative of e". h'(x) = [e"] du du will be used to find
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
![**Tutorial Exercise**
Find the derivative of the function.
\[ h(x) = e^{x^3 - x + 8} \]
---
**Step 1**
For the function \( h(x) = e^{x^3 - x + 8} \), note that \( h \) is a composite exponential function where the exponent is an expression. Hence, the generalized rule
\[
\frac{d}{dx} [e^u] = e^u \cdot \frac{du}{dx}
\]
will be used to find the derivative of \( h \).
If we think of \( e^{x^3 - x + 8} \) as \( e^u \), then \( u \) is a differentiable function of \( x \) and \( u = \)
\[ \boxed{x^3 - x + 8} \]
Apply the generalized rule to find the derivative of \( e^u \).
\[ h'(x) = \frac{d}{dx} [e^u] \]
\[ = \left(\boxed{e^u}\right) \frac{du}{dx} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd48171ba-d1cc-47b1-bf9c-8041143384f0%2Ff4457008-daed-4613-a5c0-6a51701fba27%2Fbwhrcv5_processed.png&w=3840&q=75)
Transcribed Image Text:**Tutorial Exercise**
Find the derivative of the function.
\[ h(x) = e^{x^3 - x + 8} \]
---
**Step 1**
For the function \( h(x) = e^{x^3 - x + 8} \), note that \( h \) is a composite exponential function where the exponent is an expression. Hence, the generalized rule
\[
\frac{d}{dx} [e^u] = e^u \cdot \frac{du}{dx}
\]
will be used to find the derivative of \( h \).
If we think of \( e^{x^3 - x + 8} \) as \( e^u \), then \( u \) is a differentiable function of \( x \) and \( u = \)
\[ \boxed{x^3 - x + 8} \]
Apply the generalized rule to find the derivative of \( e^u \).
\[ h'(x) = \frac{d}{dx} [e^u] \]
\[ = \left(\boxed{e^u}\right) \frac{du}{dx} \]
Expert Solution
Step 1
I ha e used chain rule of derivative as mentioned in question.
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