The problem of derivatives in the English version of the English version is as follows: let me under

1.Find the angle of inclination of the curve y=(x+3) (x-3) tangent at point (2,5).

y'=[(x+3)'(x-3)-(x-3)'(x+3)]/(x-3)²= -6/(x-3)² y'∣x=2= -6

The slope of the tangent k= -6 tan = -6 The angle of inclination of the tangent is +arctan(-6).

2.Which point of the curve y=x (x-1) is the tangent parallel to the straight line x+y+5=0?

x+y+5=0, y= -x+5 so k= -1

y'= [x'(x-1)-x(x-1)']/(x-1)²= -1/(x-1)² y'=-1/(x-1)²= -1

Hence x=2 or x=0

When x=2 , y=2

When x=0, y=0

So (2, 2) ,0, 0) is what you want.

3.Find the tangent equation for the curve y=(x+3)(x 2-1) at the point (2,15).

y'=(x+3)'(x²-1)+(x+3)(x²-1)'=3x²+6x-1

y'∣x=2=25

The tangent equation for the crossing point (2,15) is y-15=25*(x-2) i.e. 25x-y-45=0

1) Knowing the curve y=(x+3) (x-3), find the inclination angle of the straight line that passes through the point (2,5) and is tangent to the curve.

2) Knowing the curve y=x (x-1), the tangent of a point on the curve is parallel to the straight line x+y+5=0, and the coordinates of the point are calculated.

3) Find the equation of the straight line that has passed (2,15) and is tangent to the curve y=(x+3)(x 2-1).

(a) estimate a second point on the tangent line.

Estimate that the second point is on the tangent.

t, p) = __

b) calculate the rate of change of the function at the labeled point. (round your answer to one decimal place.)

The rate of change in the function at the marked point is calculated. (One decimal place for the answer).

thousand employees per year

Thousands of employees per year.

c) calculate the percentage rate of change of the function at the

labeled point. (round your answer to three decimal places.)

Calculate the percentage change in the marker point function. (3 decimal places).

per year.