Math  /  Calculus

QuestionWhat is the first derivative of q(x)=log5(2x2+8x45x)?q(x)=\log _{5}\left(\frac{2 x^{2}+8}{x^{4} 5^{x}}\right) ?
Select the correct answer below: q(x)=4x2x2+84x1q^{\prime}(x)=\frac{4 x}{2 x^{2}+8}-\frac{4}{x}-1 q(x)=1(2x2+8)ln54xln51q^{\prime}(x)=\frac{1}{\left(2 x^{2}+8\right) \ln 5}-\frac{4}{x \ln 5}-1 q(x)=4x(2x2+8)ln54xln51q^{\prime}(x)=\frac{4 x}{\left(2 x^{2}+8\right) \ln 5}-\frac{4}{x \ln 5}-1 q(x)=x4(5)ln52x2+8q^{\prime}(x)=\frac{x^{4}(5) \ln 5}{2 x^{2}+8}

Studdy Solution

STEP 1

1. We are given the function q(x)=log5(2x2+8x45x) q(x) = \log_{5}\left(\frac{2x^{2}+8}{x^{4}5^{x}}\right) .
2. We need to find the first derivative q(x) q'(x) .

STEP 2

1. Use the change of base formula to express the logarithm in terms of natural logarithms.
2. Apply the derivative rules for logarithmic functions.
3. Simplify the expression to find the derivative.

STEP 3

Use the change of base formula to rewrite the logarithm:
q(x)=ln(2x2+8x45x)ln5 q(x) = \frac{\ln\left(\frac{2x^{2}+8}{x^{4}5^{x}}\right)}{\ln 5}

STEP 4

Apply the derivative rules for logarithmic functions. Start by differentiating the numerator:
ln(2x2+8x45x)=ln(2x2+8)ln(x45x) \ln\left(\frac{2x^{2}+8}{x^{4}5^{x}}\right) = \ln(2x^{2}+8) - \ln(x^{4}5^{x})
Differentiate each part separately.

STEP 5

Differentiate ln(2x2+8) \ln(2x^{2}+8) :
ddx[ln(2x2+8)]=12x2+8ddx[2x2+8]=12x2+84x=4x2x2+8 \frac{d}{dx}[\ln(2x^{2}+8)] = \frac{1}{2x^{2}+8} \cdot \frac{d}{dx}[2x^{2}+8] = \frac{1}{2x^{2}+8} \cdot 4x = \frac{4x}{2x^{2}+8}

STEP 6

Differentiate ln(x45x) \ln(x^{4}5^{x}) :
ln(x45x)=ln(x4)+ln(5x)=4ln(x)+xln(5) \ln(x^{4}5^{x}) = \ln(x^{4}) + \ln(5^{x}) = 4\ln(x) + x\ln(5)
Differentiate each term:
ddx[4ln(x)]=4x \frac{d}{dx}[4\ln(x)] = \frac{4}{x}
ddx[xln(5)]=ln(5) \frac{d}{dx}[x\ln(5)] = \ln(5)
Combine these results:
ddx[ln(x45x)]=4x+ln(5) \frac{d}{dx}[\ln(x^{4}5^{x})] = \frac{4}{x} + \ln(5)

STEP 7

Combine the derivatives using the quotient rule for logarithms:
ddx[ln(2x2+8x45x)]=4x2x2+8(4x+ln(5)) \frac{d}{dx}\left[\ln\left(\frac{2x^{2}+8}{x^{4}5^{x}}\right)\right] = \frac{4x}{2x^{2}+8} - \left(\frac{4}{x} + \ln(5)\right)

STEP 8

Divide by ln5\ln 5 to account for the change of base:
q(x)=1ln5(4x2x2+84xln(5)) q'(x) = \frac{1}{\ln 5} \left( \frac{4x}{2x^{2}+8} - \frac{4}{x} - \ln(5) \right)
Simplify:
q(x)=4x(2x2+8)ln54xln51 q'(x) = \frac{4x}{(2x^{2}+8)\ln 5} - \frac{4}{x\ln 5} - 1
The correct answer is:
q(x)=4x(2x2+8)ln54xln51 q'(x) = \frac{4x}{(2x^{2}+8)\ln 5} - \frac{4}{x\ln 5} - 1

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