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Question: If \[y=a{{e}^{x}}+b{{e}^{-x}}\]where a, b are the parameters then \[{{y}^{''}}=\] A) \[y\] B) \[...

If y=aex+be−xy=a{{e}^{x}}+b{{e}^{-x}}where a, b are the parameters then y′′={{y}^{''}}=
A) yy
B) y′{{y}^{'}}
C) −y′-{{y}^{'}}
D) 00

Explanation

Solution

In this particular problem, expression is given that is y=aex+be−xy=a{{e}^{x}}+b{{e}^{-x}}where a, b are the parameter here, we have to find the double derivative for that we have to first differentiate that is single derivative then you have to take derivative again by applying the rule of derivative and by further solving and get the answer.

Complete step by step solution:
In this type of problems, equation is given that is y=aex+be−xy=a{{e}^{x}}+b{{e}^{-x}} where a, b are the parameter
And here you have to find out the double derivative of this expression by applying the derivative rule.
So, in first step we have to take first derivative of given equation that is
y=aex+be−x−−−(1)y=a{{e}^{x}}+b{{e}^{-x}}---(1)
By differentiating with respect to x on equation (1) we get:
dydx=d(aex)dx+d(be−x)dx\dfrac{dy}{dx}=\dfrac{d(a{{e}^{x}})}{dx}+\dfrac{d(b{{e}^{-x}})}{dx}
By using the rule of differentiation that is if there is any constant in the derivative then make sure that constant should be taken outside.
dydx=ad(ex)dx+bd(e−x)dx\dfrac{dy}{dx}=a\dfrac{d({{e}^{x}})}{dx}+b\dfrac{d({{e}^{-x}})}{dx}
Then by applying the rule of differentiation that is d(ex)dx=ex\dfrac{d({{e}^{x}})}{dx}={{e}^{x}} and also apply the composite function of derivative which is also known as chain rule that is d(e−x)dx=d(e−x)dx×d(−x)dx\dfrac{d({{e}^{-x}})}{dx}=\dfrac{d({{e}^{-x}})}{dx}\times \dfrac{d(-x)}{dx} which can also be written as d(e−x)dx=−e−x\dfrac{d({{e}^{-x}})}{dx}=-{{e}^{-x}} substitute this above equation we get:
dydx=aex−be−x−−(2)\dfrac{dy}{dx}=a{{e}^{x}}-b{{e}^{-x}}--(2)
By again, differentiating with respect to x on equation (2) we get:
d2ydx2=d(aex)dx−d(be−x)dx\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d(a{{e}^{x}})}{dx}-\dfrac{d(b{{e}^{-x}})}{dx}
By using the rule of differentiation that is if there is any constant in the derivative then make sure that constant should be taken outside.
d2ydx2=ad(ex)dx−bd(e−x)dx\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a\dfrac{d({{e}^{x}})}{dx}-b\dfrac{d({{e}^{-x}})}{dx}
Again by applying the rule of differentiation that is d(ex)dx=ex\dfrac{d({{e}^{x}})}{dx}={{e}^{x}} and also again apply the composite function of derivative which is also known as chain rule that is d(e−x)dx=d(e−x)dx×d(−x)dx\dfrac{d({{e}^{-x}})}{dx}=\dfrac{d({{e}^{-x}})}{dx}\times \dfrac{d(-x)}{dx} which can also be written as d(e−x)dx=−e−x\dfrac{d({{e}^{-x}})}{dx}=-{{e}^{-x}} substitute this above equation we get:
d2ydx2=aex−b(−e−x)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a{{e}^{x}}-b(-{{e}^{-x}})
By simplifying further we get:
d2ydx2=aex+be−x−−−(3)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=a{{e}^{x}}+b{{e}^{-x}}---(3)
By substituting the value of equation (1) on equation (3) that is
d2ydx2=y\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=y
As we know that y′′=d2ydx2{{y}^{''}}=\dfrac{{{d}^{2}}y}{d{{x}^{2}}} substitute in above equation we get:
y′′=y{{y}^{''}}=y
Therefore, the correct option is option (A).

Note:
Here, we have asked the second derivative for that we need to find the first derivative then we need to differentiate the obtained first derivative. If the question is asked to find the third derivative then we have to differentiate the second derivative and so on. Here, we use the composite function derivative in the second term of the equation of y that means differentiate the 'outside' function, and then multiply by the derivative of the 'inside' function.