\(\displaystyle f(x)= \left(9x^4+4x^3+\frac{x}{2}\right)^9\)

using the chain rule:

 \(\displaystyle \frac{d}{dx}(f(g(x)))=f{}'(g(x))g{}'(x)\)

\(\displaystyle f{}'(x)=9(9x^4+4x^3+x/2)^8(36x^3+12x^2+1/2)\)

multiply the constant 9 into the second function to simplify answer

\(\displaystyle f(x)=\log_{9}x^2\)

using the logarithm identities change the equation to base 10: 

\(\displaystyle f(x)=\frac{\ln x^2}{\ln 9}\)

using lograthim identities simplify the numerator:

\(\displaystyle f(x)=\frac{2\ln x}{\ln 9}\)

differentiate

\(\displaystyle f{}'(x)=\frac{2*\frac{1}{x}}{\ln 9}\)

Simplify

\(\displaystyle f{}'(x)=\frac{2}{x\ln 9}\)

\(\displaystyle f(x)= (x^2 +7x)(3x^3+\frac{x}{4})\)

Using the product rule: \(\displaystyle (fg){}'=f{}'g+fg{}'\)

\(\displaystyle f{}'(x)= (2x+7)(3x^3 + x/4) + (x^2+7x)(9x^2+1/4)\)

FOIL

\(\displaystyle f{}'(x)= 6x^4+2x^2/4+21x^3+7x/4+9x^4+x^2/4+63x^3+7x/4\)

Combine like-terms

\(\displaystyle f{}'(x)= 15x^4 + 84x^3+\frac{3x^2}{4}+\frac{14x}{4}\)

\(\displaystyle x^3y^2+sin(y)=11x-x^2y\)

Differentiate the equation

\(\displaystyle 3x^2y^2 + 2x^3y*y{}' + cos(y)*y{}' = 11-(2xy+x^2*y{}')\)

Simplify

\(\displaystyle 3x^2y^2 + 2x^3y*y{}' + cos(y)*y{}' = 11-2xy-x^2*y{}'\)

place all terms with \(\displaystyle y{}'\) on one side and the other terms on the other side

\(\displaystyle 2x^3y*y{}'+cos(y)*y{}'+x^2*y{}'=11-2xy-3x^2y^2\)

Simplify

\(\displaystyle y{}'*(2x^3y+cos(y)+x^2)=11-2xy-3x^2y^2\)

Divide and solve for \(\displaystyle y{}'\)

\(\displaystyle y{}'=\frac{11-2xy-3x^2y^2}{2x^3y+cos(y)+x^2}\)

Mean Value Theorem (MVT) =  \(\displaystyle f{}'(c)= \frac{f(b)-f(a)}{b-a}\) on \(\displaystyle \left [ a,b \right ]\)

\(\displaystyle f{}'(c)= \frac{(2(3)^4-\frac{(3)^2)}{7})- (2(1)^4-\frac{(1)^2}{7}))}{3-1}\)

\(\displaystyle f{}'(c)= 79.43\)

Attempting to evaluate directly (plug in -1 for \(\displaystyle x\)) results in the indeterminate form: 

\(\displaystyle \frac{0}{0}\)

Further analysis is required:

\(\displaystyle Lim _{x \rightarrow -1} \frac{x^{^{2}}+4x+3}{x+1}\dot{} = Lim _{x \rightarrow -1} \frac{(x+3)(x+1)}{x+1}\dot{} =Lim _{x \rightarrow -1}x+3\)

This final form can be evaluated directly:

\(\displaystyle Lim _{x \rightarrow -1} x+3 = -1 + 3=2\)

Attempting to evaluate directly (plug in 2 for \(\displaystyle x\)) results in the indeterminate form: 

\(\displaystyle \frac{0}{0}\)

Further analysis is required:

\(\displaystyle Lim _{x \rightarrow 2} \frac{x^2 +5x-14}{x-2}\dot{} = Lim _{x \rightarrow 2} \frac{(x+7)(x-2)}{x-2}\dot{} =Lim _{x \rightarrow 2}x+7\)

This final form can be evaluated directly:

\(\displaystyle Lim _{x \rightarrow 2} x+7 = 2+7=9\)

We evaluate the limit directly (plug in 3 for \(\displaystyle x\)) and obtain:

\(\displaystyle \frac{36}{0}\)

from which we determine that the the function has a vertical asymptote at this point (it goes off to positive or negative infinity). The limit Does Not Exist.

This limit can be evaluated directly.

Recall that \(\displaystyle secx = \frac{1}{cosx}\)

So: 

\(\displaystyle Lim _{x \rightarrow 0} xsecx = Lim _{x \rightarrow 0} \frac{x}{cosx}= \frac{0}{1}=0\)

First observe that \(\displaystyle xcsc(2\pi x) = \frac{x}{sin2\pi x}\)

Multiplying by \(\displaystyle \frac{2\pi x}{2\pi x}\) we obtain:

\(\displaystyle Lim _{x \rightarrow 0} \frac{2 \pi x*x}{2\pi x*sin2\pi x} = Lim _{x \rightarrow 0} \frac{2\pi x}{sin2\pi x}*\frac{x}{2\pi x}\)

Limit of product is the product of limits:

\(\displaystyle Lim _{x \rightarrow 0} \frac{2\pi x}{sin2\pi x}*\frac{x}{2\pi x}=Lim _{x \rightarrow 0} \frac{2\pi x}{sin2\pi x}*Lim _{x \rightarrow 0}\frac{x}{2\pi x}\)

From the Pre-Question Text: 

\(\displaystyle Lim _{x \rightarrow 0} \frac{sinx}{x} = 1\)

So:

\(\displaystyle Lim _{x \rightarrow 0} \frac{2\pi x}{sin2\pi x}*Lim _{x \rightarrow 0}\frac{x}{2\pi x}=1*\frac{1}{2\pi}=\frac{1}{2\pi}\)