Exercise 8 | Wiskunde op Tilburg University

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.

Antwoord 2 feedback

Wrong: The derivative of

$\;^2\!\log x$ is not

Antwoord 3 feedback

Wrong:

$\sqrt[3]{x^5}\neq x^{\frac{3}{5}}$ .

See Extra explanation: alternative notation.

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.