The slope of the tangent line to [tex]f(x)=\sqrt x[/tex] at [tex]x=36[/tex] is given by the value of the derivative at [tex]x=36[/tex]. By the power rule,
[tex]f(x)=\sqrt x=x^{1/2}\implies f'(x)=\dfrac12x^{-1/2}=\dfrac1{2\sqrt x}[/tex]
so that when [tex]x=36[/tex],
[tex]f'(36)=\dfrac1{2\sqrt{36}}=\dfrac1{12}[/tex]
Function defines a value from set.The slope of the given function, f(x)=sqrt(x) at 36,6 is [tex]\dfrac{1}{12}[/tex].
A function assigns the value of each element of one set to the other specific element of another set.
Given to us
y = f(x)=√x at 36,6
We know that in order to find the tangent of the given function we need to find the derivative of the given function, therefore,
[tex]y = f(x) = \sqrt{x}\\\\y = x^{\frac{1}{2}}\\\\[/tex]
Now, differentiate the function,
[tex]\dfrac{dy}{dx} = -\dfrac{1}{2}x^{(\frac{1}{2}-1)}\\\\f'(x) = -\dfrac{1}{2}x^{(-\frac{1}{2})}\\\\f'(x) = \dfrac{-1}{2\sqrt{x}}[/tex]
Substitute the value of the x in the given function,
[tex]\begin{aligned}Tangent &= \dfrac{-1}{2\sqrt{(36)}}\\\\ &= \dfrac{-1}{2\times 6}\\\\ &=\dfrac{1}{12}\end{aligned}[/tex]
Hence, the slope of the given function f(x)=sqrt(x) at 36,6 is [tex]\dfrac{1}{12}[/tex].
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