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Math  /  Algebra

Question b) y=ln⁡(x2⋅2x+1cos⁡4x)Ln⁡x2+ln⁡2x+1−ln⁡cos⁡4x2ln⁡x+12ln⁡2x+1−4ln⁡cos⁡x2x1x+12×2−4(=sin⁡x)22x+1−4[−sin⁡x]\begin{array}{l}\text { b) } y=\ln \left(\frac{x^{2} \cdot \sqrt{2 x+1}}{\cos ^{4} x}\right) \\ \operatorname{Ln} x^{2}+\ln \sqrt{2 x+1}-\ln \cos ^{4} x \\ 2 \ln x+\frac{1}{2} \ln 2 x+1-4 \ln \cos x \\ 2 x \frac{1}{x}+\frac{1}{2} \times 2-4(=\sin x) \\ \frac{2}{2 x}+1-4[-\sin x]\end{array}

Studdy Solution

STEP 1

Assumptions
1. The given function is y=ln⁡(x2⋅2x+1cos⁡4x) y = \ln \left(\frac{x^{2} \cdot \sqrt{2x+1}}{\cos^{4} x}\right) .
2. We need to simplify the expression using logarithmic properties.
3. We will use the properties of logarithms: - ln⁡(a⋅b)=ln⁡a+ln⁡b\ln(a \cdot b) = \ln a + \ln b - ln⁡(ab)=ln⁡a−ln⁡b\ln\left(\frac{a}{b}\right) = \ln a - \ln b - ln⁡(an)=nln⁡a\ln(a^n) = n \ln a

STEP 2

Apply the logarithmic property for division to the given function:
y=ln⁡(x2⋅2x+1cos⁡4x)=ln⁡(x2⋅2x+1)−ln⁡(cos⁡4x) y = \ln \left(\frac{x^{2} \cdot \sqrt{2x+1}}{\cos^{4} x}\right) = \ln(x^{2} \cdot \sqrt{2x+1}) - \ln(\cos^{4} x)

STEP 3

Apply the logarithmic property for multiplication to the first term:
ln⁡(x2⋅2x+1)=ln⁡(x2)+ln⁡(2x+1) \ln(x^{2} \cdot \sqrt{2x+1}) = \ln(x^{2}) + \ln(\sqrt{2x+1})

STEP 4

Apply the power rule of logarithms to each term:
ln⁡(x2)=2ln⁡x \ln(x^{2}) = 2 \ln x
ln⁡(2x+1)=ln⁡((2x+1)1/2)=12ln⁡(2x+1) \ln(\sqrt{2x+1}) = \ln((2x+1)^{1/2}) = \frac{1}{2} \ln(2x+1)
ln⁡(cos⁡4x)=4ln⁡(cos⁡x) \ln(\cos^{4} x) = 4 \ln(\cos x)

STEP 5

Substitute these simplified expressions back into the original equation:
y=2ln⁡x+12ln⁡(2x+1)−4ln⁡(cos⁡x) y = 2 \ln x + \frac{1}{2} \ln(2x+1) - 4 \ln(\cos x)

STEP 6

Differentiate y y with respect to x x . Use the derivative rules for logarithms:
ddx[2ln⁡x]=2⋅1x=2x \frac{d}{dx} [2 \ln x] = 2 \cdot \frac{1}{x} = \frac{2}{x}
ddx[12ln⁡(2x+1)]=12⋅12x+1⋅2=12x+1 \frac{d}{dx} \left[\frac{1}{2} \ln(2x+1)\right] = \frac{1}{2} \cdot \frac{1}{2x+1} \cdot 2 = \frac{1}{2x+1}
ddx[−4ln⁡(cos⁡x)]=−4⋅(−tan⁡x)=4tan⁡x \frac{d}{dx} [-4 \ln(\cos x)] = -4 \cdot \left(-\tan x\right) = 4 \tan x

STEP 7

Combine these derivatives to find the derivative of y y :
dydx=2x+12x+1+4tan⁡x \frac{dy}{dx} = \frac{2}{x} + \frac{1}{2x+1} + 4 \tan x
This is the derivative of the given function.

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