Find the equation of the tangent line to the curve y = sqrt (x+5) at the point (4,3)
dy/dx = (1/2(x+5)^(-1/2)
or
= 1/(2√(x+5) )
at (4,3)
dy/dx = 1/(2√9) = 1/6
y-3 = (1/6)(x-4)
6y-18 = x-4
x - 6y = -14
or y = (1/6)x + 7/3
Thanks! Sorry to bother you but do you know how I can get the same answer by solving using the definition of the derivative instead?
I used the derivative.
Do you want to find the derivative by "first principles" ?
Oh no that's okay
The steps you put include the chain rule version of the power rule right?
To find the equation of the tangent line to a curve at a given point, we need to find the slope of the tangent line and then use the point-slope form of a linear equation.
Let's start by finding the slope. The slope of the tangent line is equal to the derivative of the function at the given point. In this case, we need to find the derivative of the function y = √(x + 5).
To find the derivative, we can rewrite the function as y = (x + 5)^(1/2) and apply the power rule for differentiation. The power rule states that if we have a function of the form f(x) = x^n, then the derivative is given by f'(x) = nx^(n-1).
Applying the power rule to our function, we have:
dy/dx = (1/2)(x + 5)^(-1/2)
Now, let's evaluate the derivative at the point (4,3). Substitute x = 4 into the derivative equation:
dy/dx = (1/2)(4 + 5)^(-1/2)
= (1/2)(9)^(-1/2)
= (1/2)(1/3)
= 1/6
So, the slope of the tangent line at the point (4,3) is 1/6.
Now that we have the slope, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Plugging in the values, we have:
y - 3 = (1/6)(x - 4)
Simplifying the equation further, we get:
6y - 18 = x - 4
x - 6y = -14
Therefore, the equation of the tangent line to the curve y = √(x + 5) at the point (4, 3) is x - 6y = -14.