Let y = 2/x 2 and dy =?

Solution: dy = 2 [x (-2)]' dx =-4x (-1) dx =-(4/x) dx.

2. Explain that the following functions are at the specified point.

Is it continuous? If not continuous, yes

Can you make it up?

( 1)f(x)=(x2-x-2)/(x-2)； At point x=2

Solution: f (x) = (x 2-x-2)/(x-2) = (x-2) (x+1)/(x-2) = x+1.

So x=2 is its discontinuity.

2) Is f (x) = (1+xsinx)/x continuous at x=0?

Solution: Because f(0) is undefined, x→ 0lim (1+xsinx)/x.

=x→0lim[( 1/x)+sinx]=∞

∴x=0 is an infinite discontinuity of this function, and it is also an unrepairable discontinuity.

set up

y=x^(2/3)

Ask dy=

Solution: dy = (2/3) x (- 1/3) dx.

Let y = 2x 3+3x；; Find y''=?

Solution: y ′ = 6x2+3

y "= 12x

Y = ex (e-x), find y'=?

Solution: Take the logarithm of both sides and get:

lny= 1+(e-x)lnx

Take the derivative on both sides and you will get:

y '/

y=-lnx+(e-x)/x

∴y′=y[(-lnx)+(e-x)/x]=[ex^(e-x)][(-lnx)+(e-x)/x]

3. Explain whether the following piecewise functions are continuous? If it is not continuous, can it be supplemented?

( 1)f(x)= sin2x/x； (when x

f(x)= 2； (when x=0)

f(x)=(X2+2x)/x

(when x>0)

Solution: Because x→0-limf(x)=x→0-lim(sin2x/x).

=

x→0-lim(2x/x)=2

x→0+limf(x)= 1

x→0+lim(x^2+2x)/x

=

x→0+lim(x+2)=2

∴x→0limf(x)=f(0)=2

So the function is continuous at x=0.

y=ln( 1+x2)

, find y'=?

Solution: y ′ = 2x/(1+x 2)

Y = e (2x), find y''=?

Solution: y ′ = 2e (2x)

y″=4e^(2x).

Point out the discontinuity of the following functions and explain whether they can be supplemented.

f(x)=sin3x/x

Solution: ∫x→0 limb(x)= 1

x→0 LIMS in 3 x/x = x→0 LIMS in 3 x/x = 3

∴x=3

Is the complementary discontinuity of this function.

f(x)=2x- 1

(When x≤ 1

f(x)=(X2

–1)/(x-1) (when x >; 1)

Solution: when x> is at 1, f (x) = (x 2-1)/(x-1) = x+1.

So x→ 1+limf(x)= 1

x→ 1+lim(x+ 1)=2

And f (1) = 2-1=1,x→ 1-LIMF(x)= 1

x→ 1-lim(2x- 1)= 1

So x→ 1 limb(x) does not exist, that is, x= 1 is an irreducible discontinuity of this function.