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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

Sagot :

olemakpadu
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.

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