Substitute the expressions from the previous steps. Then substitute (x, y) = (1, 1) and solve for dy dx (9 + yeY) = d (9x + eX) (yey +endy dx ((1)e(1) + e(1)) dy dy dx = dy dx = 9+ ex = 9 + e(1) = 9+ e = What is the slope of the tangent line to the curve 9+ yey=9x + ex at the point (1, 1) for the equation 9 + yey = 9x + ex? dy dx

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Substitute the expressions from the previous steps. Then substitute \((x, y) = (1, 1)\) and solve for \(\frac{dy}{dx}\).

\[
\frac{d}{dx}(9 + ye^y) = \frac{d}{dx}(9x + e^x)
\]

\[
(ye^y + e^y)\frac{dy}{dx} = 9 + e^x
\]

\[
((1)e^{(1)} + e^{(1)})\frac{dy}{dx} = 9 + e^{(1)}
\]

\[
\left(\_\_\_\_\_\_\right)\frac{dy}{dx} = 9 + e
\]

\[
\frac{dy}{dx} = \_\_\_\_\_\_
\]

What is the slope of the tangent line to the curve \(9 + ye^y = 9x + e^x\) at the point \((1, 1)\) for the equation \(9 + ye^y = 9x + e^x\)?

\[
\frac{dy}{dx} = \_\_\_\_\_\_
\]
Transcribed Image Text:Substitute the expressions from the previous steps. Then substitute \((x, y) = (1, 1)\) and solve for \(\frac{dy}{dx}\). \[ \frac{d}{dx}(9 + ye^y) = \frac{d}{dx}(9x + e^x) \] \[ (ye^y + e^y)\frac{dy}{dx} = 9 + e^x \] \[ ((1)e^{(1)} + e^{(1)})\frac{dy}{dx} = 9 + e^{(1)} \] \[ \left(\_\_\_\_\_\_\right)\frac{dy}{dx} = 9 + e \] \[ \frac{dy}{dx} = \_\_\_\_\_\_ \] What is the slope of the tangent line to the curve \(9 + ye^y = 9x + e^x\) at the point \((1, 1)\) for the equation \(9 + ye^y = 9x + e^x\)? \[ \frac{dy}{dx} = \_\_\_\_\_\_ \]
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