Find the slop of the tangent to the curve y=(10-sqrt x)(5+sqrt x) at x=25
slope**
First find the derivative as a product:
d((10-sqrt x)(5+sqrt x))/dx
=(10-sqrt(x))((1/2sqrt(x))+(5+sqrt(x))(-1/2sqrt(x))
=(10-sqrt(x)-5-sqrt(x))/2sqrt(x)
=(5-2sqrt(x))/2sqrt(x)
Evaluate dy/dx at x=25
To find the slope of the tangent to the curve at a specific point, you can take the derivative of the curve and evaluate it at that point.
First, let's find the derivative of the curve y = (10 - sqrt(x))(5 + sqrt(x)) using the product rule of differentiation.
Let u = 10 - sqrt(x) and v = 5 + sqrt(x).
Using the product rule, the derivative dy/dx can be calculated as:
dy/dx = u * (dv/dx) + v * (du/dx).
First, let's find du/dx:
du/dx = d(10 - sqrt(x))/dx
= 0 - 1/(2sqrt(x)) [since the derivative of sqrt(x) is 1/(2sqrt(x))]
= -1/(2sqrt(x)).
Next, let's find dv/dx:
dv/dx = d(5 + sqrt(x))/dx
= 0 + 1/(2sqrt(x)) [since the derivative of sqrt(x) is 1/(2sqrt(x))]
= 1/(2sqrt(x)).
Plugging these values back into the derivative formula, we have:
dy/dx = (10 - sqrt(x)) * (1/(2sqrt(x))) + (5 + sqrt(x)) * (-1/(2sqrt(x))).
Simplifying the expression, we get:
dy/dx = (10 - sqrt(x))/(2sqrt(x)) - (5 + sqrt(x))/(2sqrt(x))
= (10 - 5 - sqrt(x) + sqrt(x))/(2sqrt(x))
= 5/(2sqrt(x)).
Now, to find the slope of the tangent at x = 25, we substitute x = 25 into the derivative expression:
slope = dy/dx
= 5/(2sqrt(25))
= 5/10
= 1/2.
Therefore, the slope of the tangent to the curve y = (10 - sqrt(x))(5 + sqrt(x)) at x = 25 is 1/2.