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’
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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).}![]()

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…. [From 1]
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