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asked in Calculus | 459 views

$e^\sqrt{2}$ will be the highest value of the given expression.

$\frac{e^{\sin x}}{e^{\cos x}}$ can be written as $e^{\sin x - \cos x}$.

$e$ is a famous irrational constant that is used as base of natural logarithms, and its approximate value is $2.7183$, (which is of course greater than $1$).

So now we just have to maximize $(\sin x - \cos x)$ to find the maximum value of given expression.

Clearly $\sin x$ and $\cos x$ are differentiable functions, hence their difference $(\sin x - \cos x)$ is also differentiable.

So we can differentiate $\sin x - \cos x$ to find its maximum value.

Solution of the equation $\frac{d}{dx}(\sin x - \cos x) = 0$ will give us the points where $\sin x - \cos x$ will attain its maximum value.

On differentiating $(\sin x - \cos x)$ with respect to $x$, we get $\cos x + \sin x$.

Putting $\cos x + \sin x = 0$ and simplifying we get

$\tan x = -1$

This equation has infinitely many solutions.

One of them is $x = \frac{3\pi}{4}$$x = \frac{3\pi}{4} \text{radians or 135 degrees}$ .

$\sin \left(\frac{3\pi}{4} \right ) = \frac{1}{\sqrt2} \text{ and }\cos \left(\frac{3\pi}{4} \right ) = \frac{-1}{\sqrt2}$.

Putting these values in $e^{\sin x - \cos x}$, we get $e^{\sqrt2}$ or $e^{1.414}$.

selected

Say,Y=esinx / ecosx

=>log Y =log(esinx / ecosx)

=>log Y =log esinx - log ecosx

=sinx - cosx

=1

log Y =log e

Y=e

sin x - cos x = 1?

what is the error if x=⊼/2

then sin⊼/2 -cos⊼/2=1 . isnot it?

yes, but that need not be maximum as cos x can go negative. $\frac {1}{\sqrt 2} - (-\frac{1}{\sqrt 2})$ gives the maximum value here.