Sorry - I can't change the Title, I put something random in whilst checking my Latex syntax now I can't change it to something more appropriate could a mod please do it for me? Thanks

I've copied out the next question and my attempt at an answer, could you please advise as to what I have got wrong;

a)In the each of the following parts, you should simplify your answers where it is appropriate to do so.

a)

i) Write down the derivative of each of the functions

$\displaystyle f(x)=x^6$ and $\displaystyle g(x)=e^{4x}$

ii) Hence, by using the Product Rule, differentiate the funtion

$\displaystyle k(x)=x^6 e^{4x}$

b)

i) Write down the derivative of each of the functions

$\displaystyle f(t)=8-t^3$ and $\displaystyle g(t)=\cos(3t)$

ii) Hence, by using the Quotient Rule, differentiate the funtion

$\displaystyle k(t)=\frac{8-t^3}{cos(3t)}$ $\displaystyle (-\frac{1}{6}\pi<t<\frac{1}{6}\pi)$

c)

i) Write down the derivative of the function

$\displaystyle f(x)=5 \ln(2x)$ $\displaystyle (x>0)$

ii) Hence, by using the Product Rule, differentiate the funtion

$\displaystyle k(x)=\sin(5 \ln(2x))$ $\displaystyle (x>0)$

i) The derivatives are;

$\displaystyle f(x)=x^6$

$\displaystyle f'(x)=6x^5$

$\displaystyle g(x)=e^{4x}$

$\displaystyle g'(x)=e^{4x}$

ii) Differentiate the function using the product rule

$\displaystyle k(x)=x^6 e^{4x}$

$\displaystyle k'(x)=(6x^5)(e^{4x}) + (x^6)(e^{4x})$

b)

i) The derivatives are;

$\displaystyle f(t)=8-t^3$

$\displaystyle f'(t)=-3t^2$

$\displaystyle g(t)=\cos(3t)$

$\displaystyle g'(t)=-\sin(3t)$

ii)Differentiate the function using the Quotient rule

$\displaystyle k(t)=\frac{8-t^3}{cos(3t)}$ $\displaystyle (-\frac{1}{6}\pi<t<\frac{1}{6}\pi)$

$\displaystyle k'(t)\frac{\cos(3t)(-3t^2)-(8-t^3)(\sin(3t))}{\cos(3t)^2}$

c)

i) The derivatives are;

$\displaystyle f(x)=5 \ln(2x)$ $\displaystyle (x>0)$

$\displaystyle f'(x)=\frac{1}{x} 5(2x)$

ii)Differentiate the function using the Composite Rule

$\displaystyle k(x)=\sin(5 \ln(2x))$ $\displaystyle (x>0)$

$\displaystyle k(x)=\sin(5 \ln(2x)) \frac{1}{x} 5(2x)$

Thank you