MHB Find the equation of the tangent line pt 3 (Need someone to check my work)

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The discussion focuses on finding the equation of the tangent line for the function y = e^(3x + cos x) at x = 0. The calculated point on the curve at x = 0 is y1 = e. The derivative of the function is y' = e^(3x + cos x) * (3 - sin x), leading to a slope (m) of 3e at that point. The equation of the tangent line is derived as y - e = 3e(x - 0). The participants confirm the calculations are correct, alleviating concerns about the value of e.
shamieh
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find the equation of the tangent line.$$y = e^{3x + cos x}$$ @ x = 0

$$
y1 = e^{3(0) + cos(0)} = e^1 = e$$
$$y1 = e$$

$$y = e^{3x+cos x}$$
$$y' = e^{3x + cos x} * (3 - sinx)$$

$$m = 3e$$

$$
y - e = 3e(x - 0) $$
 
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shamieh said:
find the equation of the tangent line.$$y = e^{3x + cos x}$$ @ x = 0

$$
y1 = e^{3(0) + cos(0)} = e^1 = e$$
$$y1 = e$$

$$y = e^{3x+cos x}$$
$$y' = e^{3x + cos x} * (3 - sinx)$$

$$m = 3e$$

$$
y - e = 3e(x - 0) $$
Looks good to me!

-Dan
 
Thank God! Was so worried about the $$e^1 = e $$
 

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