Question

maxima minima of 2^(sinx)/2^(cosx)

Answer 1

Considering Y=2^sinx-cosx and taking both side log and then differentiating function

logy=(sinx-cosx)log2

dy/dx=ylog2(cosx+sinx)

dy/dx=2^sinx-cosxlog2(cosx+sinx)

for maxima or minima dy/dx=0

as exponential function cant be zero so neglecting 2^sinx-cosx and constant(log2) we get

cosx+sinx=0

The maximum and minimum value of (a cosx + b sinx) is relatively easy to find +/-root(a²+b²)

so x=+/-root2

we get maximum and minimum values as 2^{-sin(root2)-cos(root2)} and 2^{sin(root2)-cos(root2)}

Comments

your approach is perfect but answer is :

maxima = 2^√2

^{minima = 2-√2}    this -√2 is power of 2

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and here you are finding maximum aand minimum values of sin x + cos x,,instead we have to find zeros of this equation..i am bit confused

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Max value is at pi/4

Min value is at 7pi/4

Is it correct i am getting these values

???