Maths-
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Easy

Question

If f left parenthesis x comma y right parenthesis equals S i n invisible function application open parentheses e to the power of a x end exponent plus e to the power of b y end exponent close parentheses then f subscript x y end subscript equals

  1. a b e to the power of a x plus b y end exponent times S i n invisible function application open parentheses e to the power of a x end exponent plus e to the power of b y end exponent close parentheses    
  2. negative a b e to the power of a x plus b y end exponent S i n invisible function application open parentheses e to the power of a x end exponent plus e to the power of b y end exponent close parentheses    
  3. a b e to the power of a x plus b y end exponent C o s invisible function application open parentheses e to the power of a x end exponent plus e to the power of b y end exponent close parentheses    
  4. negative a b e to the power of a x plus b y end exponent C o s invisible function application open parentheses e to the power of a x end exponent plus e to the power of b y end exponent close parentheses    

Hint:

We are given a function. It is function of two variables x and y. We should to find the value of fxy.

The correct answer is: negative a b e to the power of a x plus b y end exponent S i n invisible function application open parentheses e to the power of a x end exponent plus e to the power of b y end exponent close parentheses


    The given function is f(x,y) = sin(eax + eby)
    We have to find the value of fxy
    It means we have to find the value of fraction numerator partial differential squared f over denominator partial differential x partial differential y end fraction
    We will first take partial derivative w.r.t to one of the variable. Then we take the partial derivative of that value w.r.t the remaining variable.
    Taking the partial derivative w.r.t x
    f open parentheses x comma y close parentheses equals sin open parentheses e to the power of a x end exponent plus e to the power of b y end exponent close parentheses
fraction numerator partial differential f over denominator partial differential x end fraction equals cos left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis space fraction numerator partial differential over denominator partial differential x end fraction left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis
space space space space space space space space space equals cos left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis left parenthesis e to the power of a x end exponent plus 0 right parenthesis fraction numerator partial differential over denominator partial differential x end fraction a x
space space space space space space space space space equals a e to the power of a x end exponent cos left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis
    We will take the derivative of this value w.r.t y
    fraction numerator partial differential f over denominator partial differential x end fraction equals a e to the power of a x end exponent cos left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis
fraction numerator partial differential squared f over denominator partial differential x partial differential y end fraction space equals a e to the power of a x end exponent fraction numerator partial differential over denominator partial differential x end fraction cos left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis
space space space space space space space space space space space space space space space equals a e to the power of a x end exponent left square bracket negative sin left parenthesis e to the power of a x end exponent plus e to the power of b x end exponent right parenthesis right square bracket fraction numerator partial differential over denominator partial differential x end fraction e to the power of a x end exponent plus e to the power of b y space end exponent
space space space space space space space space space space space space space space space equals negative a e to the power of a x end exponent sin left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis left parenthesis 0 plus e to the power of b y end exponent right parenthesis fraction numerator partial differential over denominator partial differential x end fraction b y space space space space space space space space space... left curly bracket fraction numerator d over denominator d x end fraction e to the power of f left parenthesis x right parenthesis end exponent equals e to the power of f left parenthesis x right parenthesis end exponent fraction numerator d f left parenthesis x right parenthesis over denominator d x end fraction right curly bracket
space space space space space space space space space space space space space space space equals negative a e to the power of a x end exponent sin left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis left parenthesis e to the power of b y end exponent right parenthesis b
space space space space space space space space space space space space f subscript x y end subscript space space equals negative a b e to the power of a x plus b y end exponent sin left parenthesis e to the power of a x end exponent plus e to the power of b y end exponent right parenthesis space space space space space space space space space space space space space space space space... left curly bracket e to the power of m e to the power of n equals e to the power of m plus n end exponent right curly bracket
space space space space space space space space space
    This is the required answer.

    We have used rules of indices and different formulae of differentiation to solve the question. When we find the derivative w.r.t to some variable we take another variable as consta.

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