What is the tangent of y=sqrt(x)(25.5) ?

The tangent of y=sqrt(x)(25.5) is y=(12.75}{sqrt(a_{0)}}x+\frac{51sqrt(a_{0)})/4

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tangent of y=sqrt(x)(25.5)

tangent of

y = x (25.5)

y = \frac{12.75}{a _{0}} x + \frac{51 a _{0}}{4}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0} \cdot 25.5)

Find the slope of

y = x 25.5 : \frac{d y}{d x} = \frac{12.75}{x}

Compute the slope of

y = x 25.5

x = a_{0} : m = \frac{12.75}{a _{0}}

Find the line with slope m=

\frac{12.75}{a _{0}}

and passing through

(a_{0}, a_{0} 25.5) : y = \frac{12.75}{a _{0}} x + \frac{51 a _{0}}{4}

y = \frac{12.75}{a _{0}} x + \frac{51 a _{0}}{4}

7 x^{2}

y = (arcsin (sec (x)))^{t a n h (x)}

\int (a x + b) d x

\frac{d}{d x} (ln (arctan (cos (x))))

What is the tangent of y=sqrt(x)(25.5) ?
The tangent of y=sqrt(x)(25.5) is y=(12.75}{sqrt(a_{0)}}x+\frac{51sqrt(a_{0)})/4