What is the tangent of sqrt(7x+4),\at a=3 ?

The tangent of sqrt(7x+4),\at a=3 is y= 7/(2sqrt(7a+4))x+sqrt(7a+4)-(7a)/(2sqrt(7a+4))

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tangent of sqrt(7x+4),\at a=3

tangent of

7 x + 4, at a = 3

y = \frac{7}{2 7 a + 4} x + 7 a + 4 - \frac{7 a}{2 7 a + 4}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, 7 a + 4)

Find the slope of

y = 7 x + 4 : \frac{d y}{d x} = \frac{7}{2 7 x + 4}

Compute the slope of

y = 7 x + 4

x = a : m = \frac{7}{2 7 a + 4}

Find the line with slope m=

\frac{7}{2 7 a + 4}

and passing through

(a, 7 a + 4) : y = \frac{7}{2 7 a + 4} x + 7 a + 4 - \frac{7 a}{2 7 a + 4}

y = \frac{7}{2 7 a + 4} x + 7 a + 4 - \frac{7 a}{2 7 a + 4}

\frac{d}{d θ} (ln (θ))

\int \frac{3}{x ^{2} - x - 2} d x

\int ln (3 x + 4) d x

\frac{d y}{d x} = \frac{y ^{2}}{x - 1}

\frac{\partial}{\partial y} (x^{2} - 3 x y + y^{2})

What is the tangent of sqrt(7x+4),\at a=3 ?
The tangent of sqrt(7x+4),\at a=3 is y= 7/(2sqrt(7a+4))x+sqrt(7a+4)-(7a)/(2sqrt(7a+4))