Can anyone please help me with my calc summer assignment?

Using the fact that cot(x)=cos(x)/sin(x) and that cos(-pi/2)=0 and sin (-p/2) = 1 we can determine that f(-pi/2)=cot^2(-pi/2) is 0.

2) Rewrite ln (1 over the squared root of 1-x^2) using laws of logs

ln(1/(sqrt(1-x^2))

Quotient properties of logs gives

ln(1)-ln(sqrt(1-x^2))

Since ln(1) =0 this becomes

-ln(sqrt(1-x^2))

Using the definition of rational exponents

-ln[(1-x^2)^(1/2)]

Using the power property of logs

-(1/2)ln(1-x^2)

Factoring

-(1/2)ln[(1-x)(1+x)]

Product property of logs gives

-(1/2){ln(1-x)+ln(1-x)}

Disribute -(1/2)

-(1/2)ln(1-x)-(1/2)ln(1+x)

3) f(x)=x^2 + 3x write a simplified expression for (x+h) -f(x)

f(x+h)=(x+h)^2 + 3(x+h) therefore

f(x+h) -f(X)=(x+h)^2 + 3(x+h)-(x^2 + 3x)

= x^2-2xh+h^2+3x+3h-x^2-3x

=2xh+h^2+3h

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