Consider the function $y(x) = x^5 + x^3 + x + 1$. Determine $x'(1)$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$x'(1) = \dfrac{1}{9}$.
Antwoord 2 correct
Fout
Antwoord 3 optie
$x'(1) = \dfrac{1}{4}$.
Antwoord 3 correct
Fout
Antwoord 4 optie
$x'(1)$ cannot be determined, because we cannot find the prescription of $x(y)$.
Antwoord 4 correct
Fout
Antwoord 1 optie
$x'(1) = 1$.
Antwoord 1 feedback
Correct: Since the prescription of $x(y)$ cannot be determined, we use the fact that
$$x'(y) = \dfrac{1}{y'(x(y))} \qquad \text{hence} \qquad x'(1) = \dfrac{1}{y'(x(1))}.$$
In order to use this we have to determine $y'(x)$ and $x(1)$. The derivative of $y(x) =x^5 + x^3 + x + 1$ is
$$y'(x) = 5x^4 + 3x^2 + 1.$$
Moreover, note that $x(1)$ is the value of $x$ such that $y$ equals 1, hence we have to solve $y(x) = 1$:
$$
\begin{align*}
1&= x^5 + x^3 + x + 1\\
0 &= x^5 + x^3 + x = x(x^4 + x^2 + 1)\\
x = 0 &\mbox{ or } x^4 +x^2 + 1 = 0\\
&\phantom{\vee} ~~~ x^4 + x^2 = -1 \qquad \text{not possible}
\end{align*}
$$
Since $x^4\geq0$ and $x^2\geq0$, it also holds that $x^4+x^2\geq0$, hence $x^4 + x^2 = -1$ gives no solution. The unique solution is $x(1)=0$.
Now we can determine $x'(1)$:
$$x'(1) = \dfrac{1}{y'(x(1))} = \dfrac{1}{y'(0)} = \dfrac{1}{5\cdot0^4 + 3\cdot0^2 + 1} = \dfrac{1}{1} = 1.$$
Go on.
$$x'(y) = \dfrac{1}{y'(x(y))} \qquad \text{hence} \qquad x'(1) = \dfrac{1}{y'(x(1))}.$$
In order to use this we have to determine $y'(x)$ and $x(1)$. The derivative of $y(x) =x^5 + x^3 + x + 1$ is
$$y'(x) = 5x^4 + 3x^2 + 1.$$
Moreover, note that $x(1)$ is the value of $x$ such that $y$ equals 1, hence we have to solve $y(x) = 1$:
$$
\begin{align*}
1&= x^5 + x^3 + x + 1\\
0 &= x^5 + x^3 + x = x(x^4 + x^2 + 1)\\
x = 0 &\mbox{ or } x^4 +x^2 + 1 = 0\\
&\phantom{\vee} ~~~ x^4 + x^2 = -1 \qquad \text{not possible}
\end{align*}
$$
Since $x^4\geq0$ and $x^2\geq0$, it also holds that $x^4+x^2\geq0$, hence $x^4 + x^2 = -1$ gives no solution. The unique solution is $x(1)=0$.
Now we can determine $x'(1)$:
$$x'(1) = \dfrac{1}{y'(x(1))} = \dfrac{1}{y'(0)} = \dfrac{1}{5\cdot0^4 + 3\cdot0^2 + 1} = \dfrac{1}{1} = 1.$$
Go on.
Antwoord 2 feedback
Wrong: The denominator of the quotient is $y'(x(1))$, not $y'(1)$.
See Derivative inverse function, Example 1 and Example 2.
See Derivative inverse function, Example 1 and Example 2.
Antwoord 3 feedback
Wrong: The denominator of the quotient is $y'(x(1))$, not $y'(1)$.
See Derivative inverse function, Example 1 and Example 2.
See Derivative inverse function, Example 1 and Example 2.
Antwoord 4 feedback
Wrong: We do not need an explicit prescription of $x(y)$ to determine $x'(1)$.
See Derivative inverse function, Example 1 and Example 2.
See Derivative inverse function, Example 1 and Example 2.