What is the tangent of f(x)=x^{1/2},(25,5) ?

The tangent of f(x)=x^{1/2},(25,5) is y= 1/10 x+5/2

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tangent of f(x)=x^{1/2},(25,5)

tangent of

f (x) = x^{\frac{1}{2}}, (25, 5)

y = \frac{1}{10} x + \frac{5}{2}

Solution steps

Find the slope of

f (x) = x^{\frac{1}{2}} : \frac{df ( x )}{d x} = \frac{1}{2 x ^{\frac{1}{2}}}

Compute the slope of

f (x) = x^{\frac{1}{2}}

(25, 5) : m = \frac{1}{10}

Find the line with slope m=

\frac{1}{10}

and passing through

(25, 5) : y = \frac{1}{10} x + \frac{5}{2}

y = \frac{1}{10} x + \frac{5}{2}

\frac{d}{d x} (f (4 x^{5})) = 8 x^{4}

cos (6 t)

\frac{( 2 x + 7 )}{x}

3 y^{^{''}} + 3 y^{^{'}} - y = 0

What is the tangent of f(x)=x^{1/2},(25,5) ?
The tangent of f(x)=x^{1/2},(25,5) is y= 1/10 x+5/2