Differentiate the following w.r.t. \(x\):
\(\sqrt{e^{\sqrt{x}}},x>0\)
\(\sqrt{e^{\sqrt{x}}},x>0\)
Solution and Explanation
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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Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
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Concepts Used:
Exponential and Logarithmic Functions
Logarithmic Functions:
The inverses of exponential functions are the logarithmic functions. The exponential function is y = ax and its inverse is x = ay. The logarithmic function y = logax is derived as the equivalent to the exponential equation x = ay. y = logax only under the following conditions: x = ay, (where, a > 0, and a≠1). In totality, it is called the logarithmic function with base a.
The domain of a logarithmic function is real numbers greater than 0, and the range is real numbers. The graph of y = logax is symmetrical to the graph of y = ax w.r.t. the line y = x. This relationship is true for any of the exponential functions and their inverse.
Exponential Functions:
Exponential functions have the formation as:
f(x)=bx
where,
b = the base
x = the exponent (or power)