1. ## Finding derivatives

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;

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

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

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

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

b)
i) Write down the derivative of each of the functions

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

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

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

c)
i) Write down the derivative of the function

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

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

$k(x)=\sin(5 \ln(2x))$ $(x>0)$
a)
i) The derivatives are;

$f(x)=x^6$

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

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

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

ii) Differentiate the function using the product rule

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

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

b)
i) The derivatives are;

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

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

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

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

ii)Differentiate the function using the Quotient rule

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

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

c)
i) The derivatives are;

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

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

ii)Differentiate the function using the Composite Rule

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

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

Thank you

2. Originally Posted by RhysGM
...
ii) Differentiate the function using the product rule

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

$k'(x)=(6x^5)(e^{4x}) + (x^6)(e^{4x})= \color{blue}\bold{x^5 \cdot e^{4x}(6+x)}$

...

c)
i) The derivatives are;

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

$f'(x)=\frac{1}{\bold{\color{red}2}\color{black}x} 5(2) = \dfrac5{x}$

ii)Differentiate the function using the Composite Rule

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

$\bold{\color{blue}k'(x)=\cos(5 \ln(2x)) \frac{5}{x}}$

Thank you
...

3. ## Derivitives

Originally Posted by earboth
$k'(x)=(6x^5)(e^{4x}) + (x^6)(e^{4x})= \color{blue}\bold{x^5 \cdot e^{4x}(6+x)}$

Thank you for your corrections. But could you step me through how you simplified the above.

4. Originally Posted by RhysGM
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)
i) The derivatives are;

$f(x)=x^6$

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

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

$g'(x)=e^{4x}$
No. $g'(x)= 4e^{4x}$
Chain rule!

ii) Differentiate the function using the product rule

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

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

b)
i) The derivatives are;

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

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

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

$g'(t)=-\sin(3t)$
No, g'(t)= -3 sin(3t)
Chain rule again.

ii)Differentiate the function using the Quotient rule

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

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

c)
i) The derivatives are;

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

$f'(x)=\frac{1}{x} 5(2x)$
No. Notice that your solution is the same as 2/5, a constant.
$f'= \frac{5}{2x}\left(2\right)= \frac{5}{x}$
Notice that the "2" has disappeared here. That is because 5ln(2x)= 5(ln(x)+ ln(2))= 5ln(x)+ 5ln(2), and the derivative of the constant 5ln(x) is 0.

ii)Differentiate the function using the Composite Rule
I would say "chain rule"

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

$k(x)=\sin(5 \ln(2x)) \frac{1}{x} 5(2x)$
$k'(x)= \frac{5 cos(5ln(2x))}{x}$
The derivative of sin(x) is cos(x), not sin(x).

Thank you

5. I think I've figured it out;

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

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

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

6. Originally Posted by HallsofIvy
...

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

...
Originally Posted by RhysGM
I think I've figured it out; Unfortunately no. See above HallsofIvy's post

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

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

The common factor is $x^5 \cdot e^{4x}$ which I put before the bracket.