Consider the function $Z(x) = z(x,y(x))$, where
$$z(x,y) = 2^x\ln(y) \quad \text{and} \quad y(x) = x^4 + x^2 + 1.$$
Determine $Z'(2)$.
$Z'(2) =\ln(16)\ln(21) + \dfrac{48}{7}.$
$Z'(2) =4\ln(2)\ln(21) + \dfrac{4}{21}.$
$Z'(2) =4\ln(21) + \dfrac{48}{7}.$
$Z'(2) =4\ln(23) + \dfrac{48}{7}.$
Wrong: $\ln(2)\ln(y) \neq \ln(2+y).$
Try again.
Correct: The partial derivatives of $z(x,y)$ at $(x,y(x))$ and the derivative of $y(x)$ are:
$$
\begin{align*}
z'_x(x,y) &= \ln(y) \cdot 2^x \cdot \ln(2) = 2^x\ln(2)\ln(y)\\
z'_x(x,y(x)) &= 2^x\ln(2) \cdot \ln(x^4 + x^2 + 1) = 2^x\ln(2)\ln(x^4 + x^2 + 1) \\[2mm]
z'_y(x,y) &= 2^x \cdot \dfrac{1}{y} = \dfrac{2^x}{y}\\
z'_y(x,y(x)) &= \dfrac{2^x}{x^4+x^2+1}\\[2mm]
y'(x) &= 4x^3 + 2x.
\end{align*}
$$
According to the chain rule (case 2):
$$
\begin{align*}
Z'(x) &= 2^x\ln(2)\ln(x^4 + x^2 + 1) + \dfrac{2^x}{x^4+x^2+1} \cdot (4x^3 + 2x) = 2^x\ln(2)\ln(x^4 + x^2 + 1)+ \dfrac{2^x(4x^3 + 2x)}{x^4+x^2+1}.
\end{align*}
$$
Finally, we plug in $x=2$:
$$
\begin{align*}
Z'(2) &= 2^2\ln(2)\ln(2^4 + 2^2 + 1)+ \dfrac{2^2(4\cdot2^3 + 2\cdot 2)}{2^4+2^2+1} = 4\ln(2)\ln(21) + \dfrac{144}{21}\\
&= \ln(2^4)\ln(21) + \dfrac{48}{7}= \ln(16)\ln(21) + \dfrac{48}{7}.
\end{align*}
$$
Go on.
Wrong: Do not forget to apply the chain rule: special case 2.
See Chain rule (case 2), Example 1 and Example 2.
Wrong: $z'_x(x,y) \neq 2^x\ln(y)$.
See Derivatives elemenatary functions.