# [A-Level] Derivative of Transcendental Function

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#1
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:

Iโm not sure where Iโve gone wrong or If Iโm missing something, the calculator is working in degrees for both.

Iโm starting A-Levels in September so please be light with me when replying.

Thanks
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1 year ago
#2
(Original post by 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:

Iโm not sure where Iโve gone wrong or If Iโm missing something, the calculator is working in degrees for both.

Iโm starting A-Levels in September so please be light with me when replying.

Thanks
The derivative looks correct to me.
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#3
(Original post by MediocreSince01)
The derivative looks correct to me.
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.
0
1 year ago
#4
(Original post by 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'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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#5
(Original post by 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)
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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1 year ago
#6
(Original post by 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.
The correct value is 11.4300523.. I'm not sure why the Classwiz isn't giving you the correct answer then.
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1 year ago
#7
(Original post by 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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#8
(Original post by NotNotBatman)

Returns
0
1 year ago
#9
(Original post by 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:

Iโm not sure where Iโve gone wrong or If Iโm missing something, the calculator is working in degrees for both.

Iโm starting A-Levels in September so please be light with me when replying.

Thanks
The derivatives of trig functions are only valid in radians! Set it in radians
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1 year ago
#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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#11
(Original post by _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).
I just set my calculator into radians and computed the same thing - images posted already, why do I get a maths error?
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1 year ago
#12
(Original post by BrandonS15)

Returns
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.
Last edited by _gcx; 1 year ago
1
#13
(Original post by _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.
I see, using geogebra, the function ln(sin x) doesnโt even cross x = 5 so its unable to give a gradient at that point. Thanks
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1 year ago
#14
(Original post by BrandonS15)

Returns
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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1 year ago
#15
(Original post by 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?
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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1 year ago
#16
(Original post by _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.
Mixed words up, you want Im part to be in (-pi, pi) for analyticity, you can include the negative real line but this doesn't define an injective function, which gives us more useful properties.
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