Question

If y=e log_(e) 5x + e log_(e) 7x then dy/dx =

a) 12
b) 5
c) 7
d) e^12x​

Answer 1

Answer:

Answer is e^12x of the questions

Answer 2

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\boxed{ \red{\bf{ {e}^{ log_{e}(x)} = x}}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}x = 1 }}}$

$\boxed{ \red{\bf{\dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx}f(x)}}}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = {e}^{ log_{e}(5x)} + {e}^{ log_{e}(7x)}$

$\rm :\longmapsto\:y = 5x + 7x$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \because \: \boxed{ \pink{\bf{{e}^{ log_{e}(x)} = x}}}$

$\rm :\longmapsto\:y = 12x$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} y = \dfrac{d}{dx}12x$

$\rm :\longmapsto\:\dfrac{dy}{dx}= 12\dfrac{d}{dx}x$

$\rm :\longmapsto\:\dfrac{dy}{dx} = 12 \times 1$

$\bf\implies \:\dfrac{dy}{dx} = 12$

$\boxed{ \red{\bf{Hence, \: Option \: (a) \: is \: correct}}}$

Additional Information :-

$\boxed{ \blue{\bf{\dfrac{d}{dx}logx = \dfrac{1}{x} }}}$

$\boxed{ \blue{\bf{\dfrac{d}{dx} {e}^{x} = {e}^{x}}}}$

$\boxed{ \blue{\bf{\dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) }}}$

$\boxed{ \blue{\bf{\dfrac{d}{dx}sinx = cosx}}}$

$\boxed{ \blue{\bf{\dfrac{d}{dx}cosx = - sinx}}}$

$\boxed{ \blue{\bf{\dfrac{d}{dx}cosecx = - cosecx \: cotx}}}$

$\boxed{ \blue{\bf{\dfrac{d}{dx}secx = secx \: tanx}}}$