# Derivative in Degree

Plz read it . I'm going to draw the derivative of sine at 45 in degree .

y1=Sine 45 =0.707106 and y2=sine 45.001 =0.707119

so dy=y2-y1=0.000013

and dx=45.001-45=0.001

now dy/dx =0.012

I know this answer is approximately drawn. We know that the derivative of sinX is cosX . So the derivative of sine at 45 is equal to the simple value of cos45=0.707

(0.012 is not equal to 0.707) Thats why I have observed that practically the derivative of trigonometric functions is proved only in Radion . Why it is not proved in Degree ? For example (in Radion) the derivative of sine at (pi/4) =0.707 which is exactly the same for the simple value cos(pi/4)=0.707 thats proved it .

But in Degree the derivative of sine at 45 = 0.012 as I draw it above which is against the simple value of cos45= 0.707 .
I have calculated it by hand as well as by calculator . Plz Find the reason behind it . Akhtar

Note by Akhtar Ali
5 years, 6 months ago

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Very interesting. I'd like to, for a second, stop using the numerical approach to a derivative, and look at the graphical approach. The derivative is the slope of the tangent line of that curve at a specific x-value. If you consider the sine graph in degrees, instead of radians, it no longer has a period of about 6.28, but a much larger 360. Note however that the amplitude is the same at 2. If you can picture the graphs in your head, you can easily see that at x=45, the degree sine curve's slope is much smaller than the radian sine curve's slope at x=pi/4. This is the error.

For extra confirmation, think of it this way: we know that the sine function (graph) normally works in radians. So if we want the graph to be in degrees, we must alter the period. Our correct sine function (still in radians) would be $$sin(\frac{2\pi}{360} x)$$, and it would look identical to the degree sine function. Then, the derivative of this is $$\frac{2\pi}{360} cos(\frac{2\pi}{360} x)$$. Plugging in x=45 into the derivative will give the number (.012...) that you were approximating before.

- 5 years, 6 months ago

The derivative holds only radian measure.

$f(x) = \sin x \Rightarrow f' (x) = \cos x.$

If you want to convert to degree measure, then we have

$f(x) = \sin x^\circ \Rightarrow f' (x) = \frac{2\pi} { 360 } \cos x^\circ.$

The reason for this factor is the conversion from degrees to radians, namely $$x^ \circ = \frac{2 \pi}{360} x$$ radians. The constant then falls out by applying the chain rule.

A similar example is to consider the graph of $$y = x^2$$ which should have a derivative of $$2x$$. In the graph $$y = (\pi x)^2$$, if we use a substitution of $$\pi x = z$$ to get $$y = z^2$$, does it have a derivative of $$y' = 2z = 2\pi x$$? You should recognize that you're missing the term $$\frac{dz}{dx}$$ from the chain rule.

Staff - 5 years, 6 months ago

Thanks. I don't really know the proper or technical terms, as I have waited to study calculus until this coming year (I wanted to be able to learn something in math class, regularly at least). Although, I do very much understand what you are saying.

- 5 years, 6 months ago

My question is only that if the numeric values are proved in radian for derivative with the same method then why the derivative is not proved in degree.Lets again I'm going to draw the derivative of sine at (pi /4) in radian . Let y=sin(pi/4) y1=Sine (pi /4) =0.707106 and y2=sine(pi /4+0.001) =0.7078135343

so dy=y2-y1=0.00070675311

and dx=0.001

now dy/dx =0.70675311 this result is approximately true . But you can see the result was not true for degree as I solve in my first discussion . So find the reason only for numeric value only ? Thats the challenge !

- 5 years, 6 months ago

I showed you. It makes sense graphically, but it also works when you plug in the numbers.

- 5 years, 6 months ago

how it works ? can you give me any example of both cases ?

- 5 years, 6 months ago

All of you thinking that derivative can be calculated in radian or degree but I want to clear all of you that derivative is a normal number that only shows the rate of change of a function at any independent variable . It is not the value of radian or degree.

- 5 years, 6 months ago

when we use angle in radian then we follow radian system to draw the derivative and same for the degree . but the answer of derivative should be same for both cases . because the derivative is a normal number that only shows the rate of change of a function at any independent variable . It is not the value of radian or degree.

- 5 years, 6 months ago

I am totally disagree with this one f(x)=sinx∘⇒f′(x)=(2π/360)cosx∘ to convert to degree measure.

- 5 years, 6 months ago

I think and believe that numeric value is the base of every thing in math !

- 5 years, 6 months ago

What about geometry? The Greeks hardly even knew what numbers were.

- 5 years, 6 months ago

d (sin x) / dx = cos x, x is in radians.

suppose that w is the equivalent of x in degrees.

x = (2 pi / 360) w.

sin x = sin ((2 pi / 360) w).

using the chain rule, d (sin ((2 pi / 360) w)) / dw = (2 pi / 360) cos ((2 pi / 360) w).

- 5 years, 6 months ago

Its mean if we want the derivative for w=0 in degree the answer would be 0.0174 . Thats not true one . It should be one at zero degree .

- 5 years, 6 months ago

Are you sure that it is one at zero degree?(for the sin(x) graph of course) Clearly you got somewhere in your understanding wrong. This is so as when we differentiate sin(x degrees) with respect to x it WILL not give you cos(x degrees).

- 5 years, 6 months ago

Do you find any mistake in this process : the derivative of sine at 45 in degree .

y1=Sine 45 =0.707106 and y2=sine 45.001 =0.707119

so dy=y2-y1=0.000013

and dx=45.001-45=0.001

now dy/dx =0.012

- 5 years, 6 months ago

Nothing wrong with that as it is very extremely close to $$\frac{2\pi}{360^{\circ}} \cos(45^{\circ})$$. Note it is true that $$\frac{d \sin(x^{\circ})}{d x^{\circ}}=\cos (x^{\circ})$$ But NOT $$\frac{d \sin(x^{\circ})}{d x}=\cos (x^{\circ})$$

I am not so sure but maybe it will be clearer if i explain to you this:

The graph of $$y=\cos(x^{\circ})$$ when plotted with the $$x,y$$ axis, it is against $$x$$ not $$x^{\circ}$$. So when you plug in the values of 45 and 45.001, you are doing it for $$x$$, not $$x^{\circ}$$. To make it what you wish it to be, then you must compress the $$x$$ axis to $$x^{\circ}$$ axis.

- 5 years, 6 months ago

Didn't check your arithmetic but you are using dy and dx as change in y caused by change in x and dx as change in x. dy/dx is not a fraction genius it is derivative of y with respect to x and only for one level is it said to be that since y must be poly t s over the interval all same place for the interval ( be careful for leaps you probably don't understand since you stay away from what you don't know about ?

- 11 months, 3 weeks ago

d(log 1/e of t)/dt = -1/t 0<a<= t <=1

- 11 months, 3 weeks ago

Watch your text books when they use "local definition" for chapter, dx defined to be change in x and then dy/dx is a fraction for that chapter.

- 11 months, 3 weeks ago

Can you draw any example of derivative for sine in degree not graphically but numerically as I did ?

- 5 years, 6 months ago

I don't know why you continue to have difficulty understanding what everyone else has been trying to explain to you.

Consider the definition of derivative: We say that the derivative $$f'(a)$$ of a function $$f(x)$$ at a point $$x = a$$ is the limit $f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}.$ Now suppose we take $$f(x) = \sin x$$ where $$x$$ is measured in degrees. Then we can numerically estimate the value of $$f'(0)$$ by considering values of $$x$$ near $$a = 0$$. For example, $f'(0) \approx \frac{\sin 1^\circ - \sin 0^\circ}{1^\circ - 0^\circ} = \sin 1^\circ.$ That's clearly not equal to 1, nor would we expect that choosing even smaller values of $$x$$ would cause the limit to approach 1.

The reason why radians are a natural choice for angle measurement in calculus is analogous to why the base-$$e$$ logarithm is a natural choice of exponential base. These choices lead to functions whose rates of change (derivatives and antiderivatives) do not depend on some unit or base of measurement.

For trigonometric functions, a radian is defined as the angle subtending an arc of a circle whose arclength equals its radius. So in the case of a unit circle, the arclength subtended by one radian has a length of 1. This is a natural choice--it means that there is a unit-to-unit relationship between the numerical value of an angle, and the numerical value of any lengths or ratios of lengths. Informally speaking, that is why the derivative of trigonometric functions when measured in radians behave the way they do, and why measuring angles in degrees does not.

In fact, we can see how this relates to the computation we did above: the numerator $$\sin x - \sin a$$ is a difference of trigonometric ratios, but the denominator $$x - a$$, if measured in degrees, is a difference of angles. If angles are measured on a different numerical scale than the ratio of lengths (as is the case with degree measure), then why should you be surprised that the derivative depends on that choice of scale?

- 5 years, 6 months ago

Thank you. A great explanation.

- 5 years, 6 months ago

Oo No ! All the explanation was good but how are you finding the limiting value of derivative of f(x) at zero ? As x approaches to a and a=0 So now you was to find the limiting value of sinx/x as x approaches to zero but you fixed sin1 degree that's not according to the definition of derivative.
This is wrong one "f′(0)≈lim x→a(sin1∘−sin0∘)/(1∘−0∘=sin1∘)." You have to explain it !

- 5 years, 6 months ago

My definition and calculation are correct. Your understanding is incorrect. Refer to any introductory text on real analysis, such as Walter Rudin's "Principles of Mathematical Analysis."

I specifically stated that a NUMERICAL ESTIMATE (i.e., an APPROXIMATION) for the derivative can be made by choosing a suitably close value of $$x$$ to zero. The mathematical symbol for "approximately" is $$\approx$$. If that approximation is not good enough for you, then feel free to choose smaller values: \begin{align*} \sin 1^\circ &\approx 0.017452406437283512819 \\ \sin 0.1^\circ &\approx 0.0017453283658983088358 \\ \sin 0.001^\circ &\approx 0.00017453292431333680334 \\ \sin 0.0001^\circ &\approx 0.000017453292519057199614 \end{align*} and correspondingly, $f'(0) = \frac{\pi}{180} \approx 0.017453292519943295769.$ In any case, the value of the derivative when the angle is measured in degrees is not 1, as you claimed earlier.

It is presumptuous to assume that others are wrong simply because your understanding of the subject is different.

- 5 years, 6 months ago

Really Good One ! Finally I'm cleared by you . Thank You very much and Sorry to Every one whom I can't understand early . I'm very much happy now .

- 5 years, 6 months ago

The derivative is a normal number that only shows the rate of change of a function at any independent variable . It is not the value of radian or degree.

- 5 years, 6 months ago

It's " difference quotient ". of Indefinite Integral not of what it's dy/dx of . simple eg 2t = { {t+z}^2 -z{t +z} -t^2 +zt} /z lim as z>0 is t^2

- 11 months, 3 weeks ago

Stop this you idiots. derivative of sin of angle with RESPECT TO THE ANGLE is 2 times Pi times cosine of the angle . As I have told you since Reagan was president. Sin, Cosine and Angle do not depend on how you name your angles idiots . You don't understand ? Then get your asses somewhere else and quit polluting our western ways. .

- 11 months, 3 weeks ago

And there are no 'negative numbers and no negative exponents. And Log( 1/2) WRONG_ INVALID ditto for Log to base 1/2 of 3

- 11 months, 3 weeks ago