Practicing Success
Practicing Success
CUET Preparation Today
| What is $\frac{dy}{dx}$ where $y=e^{e^{e^{x^2}}}$? |
$e^{e^{e^{x^2}}}$ $e^{e^{e^{x^2}}}2x$ $e^{e^{e^{x^2}}}e^{e^{x^2}}e^{x^2}2x$ $e^{e^{e^{x^2}}}e^{x^2}$ |
| $e^{e^{e^{x^2}}}e^{e^{x^2}}e^{x^2}2x$ |
| Taking logarithm in both sides $\log y=e^{e^{x^2}}$. Taking log again we get $\log(\log y)=e^{x^2}$. Differentiating both sides w.r.to x we get $1/\log y\frac{d}{dx}(\log y)=\frac{d}{dx}(e^{x^2})$. So $\frac{1}{y\log y}\frac{dy}{dx}=e^{x^2}2x$. So$\frac{dy}{dx}=y\log ye^{x^2}2x=e^{e^{e^{x^2}}}e^{e^{x^2}}e^{x^2}2x$ |
