Determine the derivative of $y(x)=e^{x+\ln(x)}$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$y'(x)=e^{x+\ln(x)}$
Antwoord 2 correct
Fout
Antwoord 3 optie
$y'(x)=e^{1+\frac{1}{x}}$
Antwoord 3 correct
Fout
Antwoord 4 optie
None of the other answers is correct.
Antwoord 4 correct
Fout
Antwoord 1 optie
$y'(x)=(1+x)e^x$
Antwoord 1 feedback
Correct: $$\begin{align*}
y'(x) & = e^{x+\ln(x)}\cdot(1+\frac{1}{x})\\
& = e^x\cdot e^{\ln{x}}\cdot(1+\frac{1}{x})\\
& = e^x \cdot x \cdot (1+\frac{1}{x})\\
& = e^x (1+x).
\end{align*}$$
Go on.
y'(x) & = e^{x+\ln(x)}\cdot(1+\frac{1}{x})\\
& = e^x\cdot e^{\ln{x}}\cdot(1+\frac{1}{x})\\
& = e^x \cdot x \cdot (1+\frac{1}{x})\\
& = e^x (1+x).
\end{align*}$$
Go on.
Antwoord 2 feedback
Antwoord 3 feedback
Wrong: The chain rule does not state the following.
Let $y(x) = u(v(x))$ be a composite function. Then:
$$ y'(x) = u'(v'(x)).$$
See Chain rule.
Let $y(x) = u(v(x))$ be a composite function. Then:
$$ y'(x) = u'(v'(x)).$$
See Chain rule.
Antwoord 4 feedback
Wrong: The correct answer is among them.
Try again.
Try again.