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
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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