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

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SUMMARY

The equation of the tangent line for the function \(y = e^{3x + \cos x}\) at \(x = 0\) is derived as follows: First, the function evaluates to \(y(0) = e\). The derivative, calculated as \(y' = e^{3x + \cos x} \cdot (3 - \sin x)\), gives the slope at \(x = 0\) as \(m = 3e\). Thus, the equation of the tangent line is \(y - e = 3e(x - 0)\), confirming the calculations are correct.

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  • Familiarity with exponential functions and trigonometric functions
  • Knowledge of the chain rule in differentiation
  • Ability to manipulate equations to find tangent lines
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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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