If \(y=x^{e^x}+x^{e} for x>0\),the \( \dfrac{dy}{dx}\) is equal to ?
\(x^{e^x}[1/x+lnx]+e^x\)
\(x^{e^x} e^x[1/x+lnx]+ex^e-1\)
\(e^x.x^{e^x-1}+ex^e\)
\(x^{e^x }e^-x[1/x-lnx]+ex^e-1\)
\(x^{e^x} e^x[1/x-lnx]+ex^e-1\)
\(=e^{x}.x^{e^x-1}+e.x^{e-1}\)
The Correct Option is
Solution and Explanation
\(y=x^{e^x}+x^{e} \text{ for } x>0\)
Now \(\dfrac{dy}{dx}\) for \( x>0\)can be represented as
\(\dfrac{d}{dx}(x^{e^x}+x^{e})\)
\(=e^{x}.x^{e^x-1}+e.x^{e-1}\) (_Ans)
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Concepts Used:
Application of Derivatives
Various Applications of Derivatives-
Rate of Change of Quantities:
If some other quantity ‘y’ causes some change in a quantity of surely ‘x’, in view of the fact that an equation of the form y = f(x) gets consistently pleased, i.e, ‘y’ is a function of ‘x’ then the rate of change of ‘y’ related to ‘x’ is to be given by
\(\frac{\triangle y}{\triangle x}=\frac{y_2-y_1}{x_2-x_1}\)
This is also known to be as the Average Rate of Change.
Increasing and Decreasing Function:
Consider y = f(x) be a differentiable function (whose derivative exists at all points in the domain) in an interval x = (a,b).
- If for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≤ f(x2); then the function f(x) is known as increasing in this interval.
- Likewise, if for any two points x1 and x2 in the interval x such a manner that x1 < x2, there holds an inequality f(x1) ≥ f(x2); then the function f(x) is known as decreasing in this interval.
- The functions are commonly known as strictly increasing or decreasing functions, given the inequalities are strict: f(x1) < f(x2) for strictly increasing and f(x1) > f(x2) for strictly decreasing.
Read More: Application of Derivatives