What is the tangent of f(x)=sqrt(x+63),(1,8) ?

The tangent of f(x)=sqrt(x+63),(1,8) is y= 1/16 x+127/16

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tangent of f(x)=sqrt(x+63),(1,8)

tangent of

f (x) = x + 63, (1, 8)

y = \frac{1}{16} x + \frac{127}{16}

Solution steps

Find the slope of

f (x) = x + 63 : \frac{df ( x )}{d x} = \frac{1}{2 x + 63}

Compute the slope of

f (x) = x + 63

(1, 8) : m = \frac{1}{16}

Find the line with slope m=

\frac{1}{16}

and passing through

(1, 8) : y = \frac{1}{16} x + \frac{127}{16}

y = \frac{1}{16} x + \frac{127}{16}

\int_{0}^{1} t 1 + 4 t^{2} d t

x y^{^{'}} = y + e^{\frac{y}{x}}

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\frac{d x}{d t} = x sin (t), x (0) = 1

\int 6 tan (x) sec^{3} (x) d x

What is the tangent of f(x)=sqrt(x+63),(1,8) ?
The tangent of f(x)=sqrt(x+63),(1,8) is y= 1/16 x+127/16