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Math  /  Calculus

QuestionFind the derivative of y=e−xcos⁡2xx2+x+1 y = \frac{e^{-x} \cos^2 x}{x^2 + x + 1} using logarithmic differentiation.

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a kinda complicated fraction with ee, cosine, and xx all mixed up, using a special trick called logarithmic differentiation! Watch out! Remember the chain rule, product rule, *and* quotient rule — they're all hiding in here!
Also, don't forget the derivative of ln⁡(y)\ln(y) is y′y\frac{y'}{y}.

STEP 2

1. Take the natural logarithm of both sides.
2. Use logarithm rules to simplify.
3. Differentiate both sides.
4. Solve for y′y'.

STEP 3

Let's **take the natural logarithm** of both sides of our equation to make it easier to differentiate.
This is the key trick of logarithmic differentiation! ln⁡(y)=ln⁡(e−x⋅cos⁡2xx2+x+1) \ln(y) = \ln\left(\frac{e^{-x} \cdot \cos^2 x}{x^2 + x + 1}\right)

STEP 4

Now, we'll use our logarithm properties to **break down** this complicated expression.
Remember that ln⁡(a⋅b)=ln⁡(a)+ln⁡(b)\ln(a \cdot b) = \ln(a) + \ln(b), ln⁡(ab)=ln⁡(a)−ln⁡(b)\ln(\frac{a}{b}) = \ln(a) - \ln(b), and ln⁡(ab)=b⋅ln⁡(a)\ln(a^b) = b \cdot \ln(a). ln⁡(y)=ln⁡(e−x)+ln⁡(cos⁡2x)−ln⁡(x2+x+1) \ln(y) = \ln(e^{-x}) + \ln(\cos^2 x) - \ln(x^2 + x + 1) ln⁡(y)=−x+2⋅ln⁡(cos⁡x)−ln⁡(x2+x+1) \ln(y) = -x + 2 \cdot \ln(\cos x) - \ln(x^2 + x + 1) Look how much nicer that looks!

STEP 5

Time to **differentiate** both sides with respect to xx!
Remember the chain rule! ddxln⁡(y)=ddx(−x+2⋅ln⁡(cos⁡x)−ln⁡(x2+x+1)) \frac{d}{dx} \ln(y) = \frac{d}{dx} \left( -x + 2 \cdot \ln(\cos x) - \ln(x^2 + x + 1) \right) y′y=−1+2⋅1cos⁡x⋅(−sin⁡x)−1x2+x+1⋅(2x+1) \frac{y'}{y} = -1 + 2 \cdot \frac{1}{\cos x} \cdot (-\sin x) - \frac{1}{x^2 + x + 1} \cdot (2x + 1) y′y=−1−2⋅tan⁡x−2x+1x2+x+1 \frac{y'}{y} = -1 - 2 \cdot \tan x - \frac{2x + 1}{x^2 + x + 1}

STEP 6

Almost there!
Now we just need to **isolate** y′y' by multiplying both sides by yy. y′=y⋅(−1−2⋅tan⁡x−2x+1x2+x+1) y' = y \cdot \left( -1 - 2 \cdot \tan x - \frac{2x + 1}{x^2 + x + 1} \right)

STEP 7

Remember what yy was originally?
Let's **substitute** that back in. y′=e−x⋅cos⁡2xx2+x+1⋅(−1−2⋅tan⁡x−2x+1x2+x+1) y' = \frac{e^{-x} \cdot \cos^2 x}{x^2 + x + 1} \cdot \left( -1 - 2 \cdot \tan x - \frac{2x + 1}{x^2 + x + 1} \right)

STEP 8

So, the derivative is: y′=e−x⋅cos⁡2xx2+x+1⋅(−1−2⋅tan⁡x−2x+1x2+x+1) y' = \frac{e^{-x} \cdot \cos^2 x}{x^2 + x + 1} \cdot \left( -1 - 2 \cdot \tan x - \frac{2x + 1}{x^2 + x + 1} \right)

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