What is the tangent of f(x)=sqrt(x+6),\at x=1 ?

The tangent of f(x)=sqrt(x+6),\at x=1 is y= 1/(2sqrt(7))x+sqrt(7)-1/(2sqrt(7))

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tangent of f(x)=sqrt(x+6),\at x=1

tangent of

f (x) = x + 6, at x = 1

y = \frac{1}{2 7} x + 7 - \frac{1}{2 7}

Solution steps

Find the tangent point:

(1, 7)

Find the slope of

f (x) = x + 6 : \frac{df ( x )}{d x} = \frac{1}{2 x + 6}

Compute the slope of

f (x) = x + 6

x = 1 : m = \frac{1}{2 7}

Find the line with slope m=

\frac{1}{2 7}

and passing through

(1, 7) : y = \frac{1}{2 7} x + 7 - \frac{1}{2 7}

y = \frac{1}{2 7} x + 7 - \frac{1}{2 7}

\frac{x ^{2}}{7 + 6 x}

\frac{d}{d x} (R (x) (x)^{2} - x^{2})

\int \frac{sec ^{2} ( \frac{1}{x} )}{x ^{2}} d x

\int \frac{cot ^{5} ( β )}{sin ( β )} d β

\int \frac{x}{( 8 - x ^{2} ) ^{2}} d x

What is the tangent of f(x)=sqrt(x+6),\at x=1 ?
The tangent of f(x)=sqrt(x+6),\at x=1 is y= 1/(2sqrt(7))x+sqrt(7)-1/(2sqrt(7))