Question

Find the slope of the tangent line to the graph of the given function at the given value of x.

y=-5x^1/2+x^3/2; x=25

Answer 1

The slope is the differential of the function.

Recall, if y = x^n, (dy/dx) = nx^(n-1).

y=-5x^1/2+x^3/2; x = 25. To differentiate this, we do it for each term.

(dy/dx) = (1/2)(-5x^(1/2 -1)) + (3/2)x^(3/2-1)

= (-2.5)x^(-0.5) + 1.5x^(0.5)
= (-2.5)/(x^0.5) + 1.5x^0.5. Changing -ve power to positive.
Note x^0.5 = Square rooot (x).

(dy/dx) = (-2.5)/(x^0.5) + 1.5x^0.5. Note at x = 25.


(dy/dx), x = 25, = (-2.5)/(25^0.5) + 1.5*25^0.5
=(-2.5)/(5) + 1.5*5 = -0.5+ 7.5 = 7.

Using also the negative value of square root of 25, which is -5.
=(-2.5)/(-5) + 1.5*-5 = 0.5 - 7.5= -7.

Slope at x = 25 is plus or minus 7.
Cheers.