Question

cos(log x + e^{x}) dy/dx ज्ञात कीजिए

Answer 1

dy/dx = - ( 1/x  + eˣ)Sin(log x + eˣ)  यदि y  = cos(log x + eˣ)

Step-by-step explanation:

dy/dx ज्ञात कीजिए

y  = cos(log x + eˣ)

dy/dx  =   - Sin(log x + eˣ)  *  d(log x + eˣ)/dx

=> dy/dx  =  - Sin(log x + eˣ) * (  d(log x)/dx  + d(eˣ)/dx)

=>  dy/dx = - Sin(log x + eˣ) * ( 1/x  + eˣ)

=> dy/dx = - ( 1/x  + eˣ)Sin(log x + eˣ)

और अधिक जानें :

sin(x²+5)"

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Answer 2

Question :-

$\rm if \: y \: = \cos( \log x + {e}^{x} ) \: \: then \: find \: \frac{dy}{dx}$

Answer :-

$\to \rm \frac{dy}{dx} = - \sin( \log x + {e}^{x} ) \bigg \{ \frac{1 + x {e}^{x} }{x} \bigg \} \:$

Solution :-

We have ,

$\to \: \rm \: y \: = \cos( \log x + {e}^{x} ) \: \: \\ \\ \sf \: differentiate \: with \: respect \: to \: x \\ \\ \to \rm \frac{dy}{dx} = \frac{d}{dx} \: \cos( \log x + {e}^{x} ) \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \rm \frac{d}{dx} \cos x \: = - \sin x} \\ \\ \to \rm \frac{dy}{dx} = - \sin( \log x + {e}^{x} ) \frac{d}{dx} \{ \log x + {e}^{x} \} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \because \rm \frac{d}{dx} \log x = \frac{1}{x} \: \: and \: \frac{d}{dx} {e}^{x} = {e}^{x} } \\ \\ \to \rm \frac{dy}{dx} = - \sin( \log x + {e}^{x} ) \{ \: \frac{1}{x} \: + {e}^{x} \} \\ \\ \to \rm \frac{dy}{dx} = - \sin( \log x + {e}^{x} ) \bigg \{ \frac{1 + x {e}^{x} }{x} \bigg \}$