Step-by-step explanation:
We have,
f(x) = [tex]\sqrt{2x}[/tex]
To find, the value of f'(2) = ?
∴ f(x) = [tex]\sqrt{2x}[/tex]
⇒ f(x) = [tex]\sqrt{2}\sqrt{x}[/tex]
⇒ f '(x) = [tex]\sqrt{2}.\dfrac{1}{2\sqrt{x} }[/tex] ........ (1)
We know that,
y = [tex]\sqrt{x}[/tex] ⇒ [tex]\dfrac{dy}{dx}=\dfrac{1}{2\sqrt{x}}[/tex]
Put x = 2 in equation(1), we get
f '(2) = [tex]\sqrt{2}.\dfrac{1}{2\sqrt{2}}[/tex]
⇒ f '(2) = [tex]\dfrac{1}{2}[/tex]
∴ The value of f'(2) = [tex]\dfrac{1}{2}[/tex]
Thus, the value of f'(2) is [tex]\dfrac{1}{2}[/tex].
The result of the function f'(x) is 1/2
Given the function
f(x)=√2x
This can also be written as:
f(x) = (2x)^1/2
Taking the derivative:
f'(x) = 1/2(2x)^-1/2 * 2
f'(x) = 1/(2x)^1/2
Substituting x = 2 into the expression:
f'(2) = 1/(2(2))^1/2
f'(2) = 1/√4
f'(2) = 1/2
Hence the result of the function f'(x) is 1/2
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