Determine the derivative of $y(x)=\sqrt[3]{(\;^2\!\log x+\sqrt{x})^5}$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$y'(x)=\frac{5}{3}(\;^2\!\log x+\sqrt{x})^{\frac{2}{3}}\cdot (\dfrac{1}{x}+\dfrac{1}{2\sqrt{x}})$
Antwoord 2 correct
Fout
Antwoord 3 optie
$y'(x)=\frac{3}{5}(\;^2\!\log x+\sqrt{x})^{-\frac{2}{5}}\cdot (\dfrac{1}{x\ln(2)}+\dfrac{1}{2\sqrt{x}})$
Antwoord 3 correct
Fout
Antwoord 4 optie
$y'(x)=\frac{5}{3}(\dfrac{1}{x\ln(2)}+\dfrac{1}{2\sqrt{x}})^{\frac{2}{3}}$
Antwoord 4 correct
Fout
Antwoord 1 optie
None of the other answers is correct.
Antwoord 1 feedback
Correct: $y(x)=(\;^2\!\log x+\sqrt{x})^{\frac{5}{3}}$.
Hence, $y'(x)=\frac{5}{3}(\;^2\!\log x+\sqrt{x})^{\frac{2}{3}}\cdot (\dfrac{1}{x\ln(2)}+\dfrac{1}{2\sqrt{x}})$.
Go on.
Hence, $y'(x)=\frac{5}{3}(\;^2\!\log x+\sqrt{x})^{\frac{2}{3}}\cdot (\dfrac{1}{x\ln(2)}+\dfrac{1}{2\sqrt{x}})$.
Go on.
Antwoord 2 feedback
Wrong: The derivative of $\;^2\!\log x$ is not $\dfrac{1}{x}$.
See Derivatives of elementary functions.
See Derivatives of elementary functions.
Antwoord 3 feedback
Antwoord 4 feedback
Wrong: The composite power rule does not state the following.
Let $y(x) = \big(v(x)\big)^p$. Then:
$$ y'(x) = p\big(v'(x)\big)^{p-1}.$$
See Extra explanation: special cases.
Let $y(x) = \big(v(x)\big)^p$. Then:
$$ y'(x) = p\big(v'(x)\big)^{p-1}.$$
See Extra explanation: special cases.