Question:
Differentiate the following w.r.t. \(x\):
\(\sqrt{e^{\sqrt{x}}},x>0\)
Differentiate the following w.r.t. \(x\):
\(\sqrt{e^{\sqrt{x}}},x>0\)
\(\sqrt{e^{\sqrt{x}}},x>0\)
Updated On: Sep 11, 2023
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Solution and Explanation
The correct answer is \(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}},x>0\)
Let \(y=\sqrt{e^{\sqrt{x}}}\)
Then,\(y^2=e^{\sqrt{x}}\)
By differentiating this relationship with respect to \(x\),we obtain
\(\implies y^2=e^{\sqrt{x}}\)
\(⇒2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{d}{dx}(\sqrt{x}) \) [By applying the chain rule]
\(⇒2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{1}{2}.\frac{1}{\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4y\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{e^{\sqrt{x}}}\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}},x>0\)
Let \(y=\sqrt{e^{\sqrt{x}}}\)
Then,\(y^2=e^{\sqrt{x}}\)
By differentiating this relationship with respect to \(x\),we obtain
\(\implies y^2=e^{\sqrt{x}}\)
\(⇒2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{d}{dx}(\sqrt{x}) \) [By applying the chain rule]
\(⇒2y\frac{dy}{dx}=e^{\sqrt{x}}\frac{1}{2}.\frac{1}{\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4y\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{e^{\sqrt{x}}}\sqrt{x}}\)
\(⇒\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}},x>0\)
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Exponential Functions:
Exponential functions have the formation as:
f(x)=bx
where,
b = the base
x = the exponent (or power)