# Find an equation of the tangent line to the curve at the given point. y = √x , at point (36, 6)?

**Solution:**

y = √x , at point (36, 6)

Now we have to find the slope of your tangent by calculating the derivative of your function and substituting the value x = 36.

f(x) = √x

f’(x) = 1/(2√x)

Substitute the value x in f’(x),

f’(36) = 1/(2√36)

f’(36) = 1/(2 × 6)

f’(36) = 1/12

So, the equation of the tangent with slope m = 1/12 and passing through (x_{0} = 36, y_{0 }= 7) can be written as:

(y - y\(_0\)_{ }= m(x - x\(_0\)

y - 7 = (1/12)(x - 36)

y = (x/12) + 4

Therefore, the equation of the tangent is y = (x/12) + 4.

## Find an equation of the tangent line to the curve at the given point. y = √x , at point (36, 6)?

**Summary:**

The equation of the tangent line to the curve at the given point y = √x , at point (36, 6) is y = (x/12) + 4.

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