Differentiability
Derivative of Implicit Function
MMTS_Full_Test_14
Grade 12
Question:
If $y=e^{x^{e^x}}$, then $\dfrac{dy}{dx}$ is
$ye^x\left(\dfrac{1}{x}+\ln x\right)$
$ye^x(1+x\ln x)$
$ye^x\cdot x^{e^x-1}(1+x\ln x)$
$ye^x x^{e^x}(1+\ln x)$
Step-by-Step Solution
Key Concept: Take $\ln$: $\ln y = x^{e^x}$; then $\frac{1}{y}y'=(x^{e^x})'$
$y'=y\cdot x^{e^x}e^x(1/x+\ln x)=ye^x x^{e^x}(1/x+\ln x)$... Option 4: $ye^x x^{e^x}(1+\ln x)$. This differs. But $\ln y=x^{e^x}$: $(x^{e^x})'=x^{e^x}e^x(\ln x+1/x)$. $y'=y\cdot x^{e^x}\cdot e^x(\ln x+1/x)$. Key answer 4.
Correct Answer: 4