Q7 of 29 Page 1

If ey (x + 1) = 1, show that

Given,


ey (x + 1) = 1


….. [1]


As,


ey (x + 1) = 1


Differentiating both side with respect to ‘x’



Product rule: if f(x) and g(x) are two differentiable functions then


[f(x)g(x)] = f(x)g(x) + g(x)f(x)


Now, applying Product rule on LHS and differentiation of any constant value is 0.



Chain rule: Let F = f o g, or equivalently, F(x) = f(g(x)) for all x. then F(x) = f(g(x))g(x)


So



{displaystyle F'(x)=f'(g(x))g'(x).}


{displaystyle F'(x)=f'(g(x))g'(x).}




…. [From 1]


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