# [A-Level] Derivative of Transcendental Function

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I have this function:

h(x) = ln(sin x)

I computed the derivative through this extended method of the chain rule:

The problem is when I try finding the value of dh/dx when x = 5 I get different answers.

cos(5)/sin(5) = 11.4300523 from my calculator

But using the derivative function on my calculator I get a different output:

Iโm not sure where Iโve gone wrong or If Iโm missing something, the calculator is working in degrees for both.

Iโm starting A-Levels in September so please be light with me when replying.

Thanks

h(x) = ln(sin x)

I computed the derivative through this extended method of the chain rule:

The problem is when I try finding the value of dh/dx when x = 5 I get different answers.

cos(5)/sin(5) = 11.4300523 from my calculator

But using the derivative function on my calculator I get a different output:

Iโm not sure where Iโve gone wrong or If Iโm missing something, the calculator is working in degrees for both.

Iโm starting A-Levels in September so please be light with me when replying.

Thanks

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#2

(Original post by

I have this function:

h(x) = ln(sin x)

I computed the derivative through this extended method of the chain rule:

The problem is when I try finding the value of dh/dx when x = 5 I get different answers.

cos(5)/sin(5) = 11.4300523 from my calculator

But using the derivative function on my calculator I get a different output:

Iโm not sure where Iโve gone wrong or If Iโm missing something, the calculator is working in degrees for both.

Iโm starting A-Levels in September so please be light with me when replying.

Thanks

**BrandonS15**)I have this function:

h(x) = ln(sin x)

I computed the derivative through this extended method of the chain rule:

The problem is when I try finding the value of dh/dx when x = 5 I get different answers.

cos(5)/sin(5) = 11.4300523 from my calculator

But using the derivative function on my calculator I get a different output:

Iโm not sure where Iโve gone wrong or If Iโm missing something, the calculator is working in degrees for both.

Iโm starting A-Levels in September so please be light with me when replying.

Thanks

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(Original post by

The derivative looks correct to me.

**MediocreSince01**)The derivative looks correct to me.

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#4

(Original post by

Then why my substitution of x = 5 into the derivative computed and the calculator computing the derivative itself and giving me the output of x = 5 different? Thats the bit I'm confused on, I'm not sure if I've missed something because I did only start learning the chain rule a few hours ago.

**BrandonS15**)Then why my substitution of x = 5 into the derivative computed and the calculator computing the derivative itself and giving me the output of x = 5 different? Thats the bit I'm confused on, I'm not sure if I've missed something because I did only start learning the chain rule a few hours ago.

I know it's correct because there's a cheat way of differentiating functions of the form ln[f(x)]

The derivative is simply f'(x) divided by f(x)

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(Original post by

I'm not sure, perhaps try again by retyping - you may have accidentally missed something!

I know it's correct because there's a cheat way of differentiating functions of the form ln[f(x)]

The derivative is simply f'(x) divided by f(x)

**MediocreSince01**)I'm not sure, perhaps try again by retyping - you may have accidentally missed something!

I know it's correct because there's a cheat way of differentiating functions of the form ln[f(x)]

The derivative is simply f'(x) divided by f(x)

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#6

(Original post by

I already have tried and checked, it seems to be entered into the calculator the same and computing cos(5)/sin(5) does give the same output. The calculator is set to work in degrees, I'm still confused.

**BrandonS15**)I already have tried and checked, it seems to be entered into the calculator the same and computing cos(5)/sin(5) does give the same output. The calculator is set to work in degrees, I'm still confused.

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#7

**BrandonS15**)

I already have tried and checked, it seems to be entered into the calculator the same and computing cos(5)/sin(5) does give the same output. The calculator is set to work in degrees, I'm still confused.

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#9

**BrandonS15**)

I have this function:

h(x) = ln(sin x)

I computed the derivative through this extended method of the chain rule:

The problem is when I try finding the value of dh/dx when x = 5 I get different answers.

cos(5)/sin(5) = 11.4300523 from my calculator

But using the derivative function on my calculator I get a different output:

Iโm not sure where Iโve gone wrong or If Iโm missing something, the calculator is working in degrees for both.

Iโm starting A-Levels in September so please be light with me when replying.

Thanks

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#10

"Correct" answer is cos(5

*radians*)/(sin(5*radians*)) which is about -0.2958... Calculus is almost invariably done with arguments in radians. The derivative of ln(sin(x)) with x measured in degrees is not cot(x) but rather pi/180 cot(x).
Last edited by _gcx; 1 year ago

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(Original post by

"Correct" answer is cos(5

**_gcx**)"Correct" answer is cos(5

*radians*)/(sin(5*radians*)) which is about -0.2958... Calculus is almost invariably done with arguments in radians. The derivative of ln(sin(x)) with x measured in degrees is not cot(x) but rather pi/180 cot(x).
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#12

Last edited by _gcx; 1 year ago

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(Original post by

sin(5) is negative, and the log of a negative number gets complicated. In short the derivative doesn't exist at x=5 because we haven't defined ln for negative numbers. The correct answer is cot(5) if you do extend the logarithm (using a branch that is continuous on the positive real axis) to negative arguments but this is beyond A-level.

**_gcx**)sin(5) is negative, and the log of a negative number gets complicated. In short the derivative doesn't exist at x=5 because we haven't defined ln for negative numbers. The correct answer is cot(5) if you do extend the logarithm (using a branch that is continuous on the positive real axis) to negative arguments but this is beyond A-level.

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#14

Now that this is defined we ask what value do we get when we put x=5 and that is Log(-sin(5)) +i*pi, so now we have a function that is defined at that point and can be shown to be differentiable on its domain using cauchy riemman equations.

But this question is presented to you because you can work it out without using any non real complex numbers as the answer is a real number, its just that your calculator can't evaluate at a non real complex part or maybe it can if your calculator is in complex mode?

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#15

(Original post by

And complex mode, but I assume you haven't learnt the complex logarithm. Beyond A level there is a function called the complex logarithm which is defined everywhere on the complex plane (if you do further maths you'll have heard of this), except at 0 (and for im part k-pi, Pi] if you want continuity).

Now that this is defined we ask what value do we get when we put x=5 and that is Log(-sin(5)) +i*pi, so now we have a function that is defined at that point and can be shown to be differentiable on its domain using cauchy riemman equations.

But this question is presented to you because you can work it out without using any non real complex numbers as the answer is a real number, its just that your calculator can't evaluate at a non real complex part or maybe it can if your calculator is in complex mode?

**NotNotBatman**)And complex mode, but I assume you haven't learnt the complex logarithm. Beyond A level there is a function called the complex logarithm which is defined everywhere on the complex plane (if you do further maths you'll have heard of this), except at 0 (and for im part k-pi, Pi] if you want continuity).

Now that this is defined we ask what value do we get when we put x=5 and that is Log(-sin(5)) +i*pi, so now we have a function that is defined at that point and can be shown to be differentiable on its domain using cauchy riemman equations.

But this question is presented to you because you can work it out without using any non real complex numbers as the answer is a real number, its just that your calculator can't evaluate at a non real complex part or maybe it can if your calculator is in complex mode?

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#16

(Original post by

you'll get discontinuities at places other than zero. if you choose the imaginary part to fall in (-pi,pi] you'll get a branch cut (line of discontinuity) on the negative real axis because the imaginary part flips sign as you cross the axis. there exists a holomorphic logarithm on any connected open subset not containing 0 though.

**_gcx**)you'll get discontinuities at places other than zero. if you choose the imaginary part to fall in (-pi,pi] you'll get a branch cut (line of discontinuity) on the negative real axis because the imaginary part flips sign as you cross the axis. there exists a holomorphic logarithm on any connected open subset not containing 0 though.

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