What is the tangent of f(x)=sqrt(x+7),(9,4) ?

The tangent of f(x)=sqrt(x+7),(9,4) is y= 1/8 x+23/8

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tangent of f(x)=sqrt(x+7),(9,4)

tangent of

f (x) = x + 7, (9, 4)

y = \frac{1}{8} x + \frac{23}{8}

Solution steps

Find the slope of

f (x) = x + 7 : \frac{df ( x )}{d x} = \frac{1}{2 x + 7}

Compute the slope of

f (x) = x + 7

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

Find the line with slope m=

\frac{1}{8}

and passing through

(9, 4) : y = \frac{1}{8} x + \frac{23}{8}

y = \frac{1}{8} x + \frac{23}{8}

\frac{d x}{d t} = \frac{2 x ^{2} + t 7 t ^{2} + x ^{2}}{2 t x}

f (x) = cos (\frac{e ^{x}}{x ^{4} + 1})

7 t \cdot \frac{d y}{d t} + y = t

3 - x^{2}, 6 - 4 x

\int (1 + x) (x - 2)^{7} d x

What is the tangent of f(x)=sqrt(x+7),(9,4) ?
The tangent of f(x)=sqrt(x+7),(9,4) is y= 1/8 x+23/8