What is the tangent of y= 1/(4+3x),(1, 1/7) ?

The tangent of y= 1/(4+3x),(1, 1/7) is y=-3/49 x+10/49

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tangent of y= 1/(4+3x),(1, 1/7)

tangent of

y = \frac{1}{4 + 3 x}, (1, \frac{1}{7})

y = - \frac{3}{49} x + \frac{10}{49}

Solution steps

Find the slope of

y = \frac{1}{4 + 3 x} : \frac{d y}{d x} = - \frac{3}{( 4 + 3 x ) ^{2}}

Compute the slope of

y = \frac{1}{4 + 3 x}

(1, \frac{1}{7}) : m = - \frac{3}{49}

Find the line with slope m=

- \frac{3}{49}

and passing through

(1, \frac{1}{7}) : y = - \frac{3}{49} x + \frac{10}{49}

y = - \frac{3}{49} x + \frac{10}{49}

f^{^{'}} (x) = 2 - 4

\frac{d}{d x} (e^{x} sec (x))

\int \frac{sin ^{3} ( x )}{cos ^{3} ( x )} d x

\frac{d}{d x} (8 x^{\frac{1}{2}})

What is the tangent of y= 1/(4+3x),(1, 1/7) ?
The tangent of y= 1/(4+3x),(1, 1/7) is y=-3/49 x+10/49