Question

differentiate using logarithmic differntiation..

Answer 1

\[f(x) =\frac{\sqrt[3]{x^3 +4} (arcsec3x)}{(\csc2x) e^{5x}}\]

Answer 2

my answer is \[f(x)(\frac{x^3}{x^3+4}+\frac{3x}{arcsec(3x)(\sqrt{1-9x^2})} +2\cot(2x) -5\]

Answer 3

Do they want you to write it like this ?

\[ \ln y = \ln (^3\sqrt{x+4} \cdot \sec^{-1}(3x)) - \ln (\csc2x \cdot e^{5x}) \]

Answer 4

yes. i have my answer, just checking.

Answer 5

Assuming that the terms are positive we can distribute that out even more and then use implicit differentiation, just can't tell yet if that makes it any easier than the quotient rule (-:

Answer 6

Did you come to that point?

\[ \ln y = \ln (\sqrt[3]{x+4})+\ln(\sec^{-1}(3x))- \ln (\csc(2x))-5x \]

Answer 7

yes, i did that.

@Spacelimbus why is it -5x only at the last term?

Answer 8

I haven't carried out the remaining step yet, but keep in mind that y is a function of x, so when you differentiate that implicitly then you get

\[\frac{1}{y} \cdot \frac{dy}{dx} \] on the left hand side of the equation, well @kaiz122
the last term is

\[\ln (e^{5x}) = 5x \ln e = 5x\]

Answer 9

oh, yes. haha. sorry.

Answer 10

Do you understand the lefthand side? Because if you want dy/dx you have to multiply your entire equation times y at the end. and y again is the function described in the problem.

Answer 11

yes, :)

Answer 12

Seems to be rather painful to me (-:

Answer 13

i've got my answer above, it's f(x)*(blablabla)

Answer 14

good, lets make that step by step then because I don't have any table here at the moment, not quite sure about the derivative of the inverse function of secant.

First term

\[ \ln (\sqrt[3]{x+4}) \longrightarrow^{'} \frac{1}{\sqrt[3]{x+4}} \cdot \frac{1}{3 \sqrt[3]{x+4}^2} \]

Answer 15

ok, i get that.


\[\ln(u ^{ \frac{1}{3}}) = \frac{1}{3} \ln (u) \]

just to make it easier?

Answer 16

sure

Answer 17

after simplification you well get

\[ \frac{1}{3x+12}\] for the first term.

Answer 18

the second term is a bit more nasty to differentiate (for me) because I am not too sure about the derivative of the inverse sec function, but anyway:

\[ \ln (\sec^{-1}(3x)) \longrightarrow \frac{1}{\sec^{-1}(3x)} \cdot \frac{1}{x\sqrt{9x^2-1}}\]

Answer 19

the next term is easier

\[ \ln (\csc(2x)) \longrightarrow \frac{1}{\csc(2x)} \cdot 2 \cot(2x) \csc (2x) \\ =2 \cot (2x) \]

Answer 20

there's a minus sign missing, but for the final answer the term will be positive anyway.

Answer 21

and I think that's it, for the last term you only get 5.

Answer 22

so, i my answer is correct? thank you very much for your help. :)

Answer 23

I don't see where you get the x^3 from in the first numerator.

Answer 24

\[ f(x) \left( \frac{1}{3x+12} + \frac{1}{x\sec^{-1}(3x) \sqrt{9x^2-1}} + 2 \cot(2x) -5 \right) \]

Answer 25

maybe you have seen the problem wrong. it's \[\sqrt[3]{x^3+4}\]

Answer 26

ah yes, but then it should be a 3x^2 shouldn't it?

Answer 27

let me try it out real quick.

Answer 28

it's an x^2. maybe i typed it wrong

Answer 29

ah ok, well then your answer seems right to me yes. I dropped that entire exponent.

Answer 30

Doesn't change much on the solution, just that you have the exponent in the numerator.

Answer 31

yes yes. thank you again. :-)

Answer 32

welcome