Answer:
[tex]\frac{dy}{dx} = (lnx)^{cosx } \cdot [\frac{cosx }{ln x} \cdot \frac{1}{x} - sin x ( ln ( ln x)) ][/tex]
Step-by-step explanation:
[tex]y = (ln x)^{cosx}\\ln y = cos x (ln ( ln x)) \\[/tex] [tex][ taking \ logarithm \ on \ both \ sides ][/tex]
[tex]\frac{1}{y} \frac{dy }{dx} = \frac{cosx }{ln x} \cdot \frac{1}{x} + ln ( ln x) (-sin x)[/tex] [tex][Chain rule : uv = u \frac{dv}{dx} + v \frac{du}{dv} ][/tex]
[tex]\frac{dy}{dx} = y \times [\frac{cosx }{ln x} \cdot \frac{1}{x} - sin x ( ln ( ln x)) ][/tex]
[tex]\frac{dy}{dx} = (lnx)^{cosx } \cdot [\frac{cosx }{ln x} \cdot \frac{1}{x} - sin x ( ln ( ln x)) ][/tex] [tex][ substituting \ the \ value \ of \ y][/tex]
