find under root sin x first principle
Answers
Answer:
Step-by-step explanation:
Many of the concepts of differentiation and integration were known and understood by the ancient Greek mathematicians, but we had to wait until the time of Isaac Newton and Gottfried Leibniz for the thinking (and algebra) we use today.
To find the derivative (the "rate of change") of the function f(x), we need to find the following limit:
\lim_{h\,\to\,0}\frac{f(x+h)-f(x)}{h}
Using this, we can find derivatives of polynomial expressions like:
\frac{d}{dx}x^2=A. Derivative of sin(x) by First Principles
We need a few results before we can find this derivative.
(A1) Limit of sin θ/θ as x → 0
First, we need this common and well-known limit.
\lim_{\theta\,\to\,0}\frac{\sin\theta}{\theta}
First, we draw a graph of y = sin θ/θ and can see the limit of the function as θ approaches 0 is 1. (Of course there is a "hole" in the graph at θ = 0.)
sin theta over theta
So we have:
\lim_{\theta\,\to\,0}\frac{\sin\theta}{\theta}=1
You can see an exploration of this result here.
(A2) Difference of 2 Sines
We also need the following trigonometric identity. (You can prove this one by expanding out the sine and cosine terms on the right.)
\sin{A}-\sin{B}=2\sin\left(\frac{A-B}{2}\right)\cos\left(\frac{A+B}{2}\right)
(A3) Fraction on a fraction
This one always confuses students. One divided by half is 2.
\large{\frac{1}{\frac{1}{2}}=2}
This is true because:
\large{\frac{1}{\frac{1}{2}}=1\div\frac{1}{2}=1\times{2}=2}
On with the Derivative of sine x
We are now ready to find the derivative of sin(x) from first principles.
Setting aside the limit for now, our first step is to evaluate the fraction with f(x) = sin x.
\frac{f(x+h)-f(x)}{h}=\frac{\sin(x+h)-\sin{x}}{h}
On the right hand side we have a difference of 2 sines, so we apply the formula in (A2) above:
\sin(x+h)-\sin x=2\sin\left(\frac{(x+h)-x}{2}\right)\cos\left(\frac{(x+h)+x}{2}\right)
Simplifying the right hand side gives:
2\sin\left(\frac{h}{2}\right)\cos\left(x+\frac{h}{2}\right)
Now to put it all together and consider the limit:
\lim_{h\,\to\,0}\frac{f(x+h)-f(x)}{h}
=\lim_{h\,\to\,0}\large{\frac{2\sin\left(\frac{h}{2}\right)\cos\left(x+\frac{h}{2}\right)}{h}}
We make use of (3), fraction on a fraction, to bring that 2 out front down to the bottom:
=\lim_{h\,\to\,0}\large{\frac{\sin\left(\frac{h}{2}\right)\cos\left(x+\frac{h}{2}\right)}{\frac{h}{2}}}
Now, the limit of a product is the product of the limits, so we can write this as:
=\lim_{h\,\to\,0}\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}} \times{\lim_{h\,\to\,0}\cos\left(x+\frac{h}{2}\right)}
Now, the first limit is in the form of Limit of sin θ/θ that we met in (A1) above.
We know it has value 1.
For the right hand limit, we simply obtain cos x.
So we can conclude that
\frac{d}{dx}\sin{x}=\cos{x}
B. Derivative of square root of sin x from first principles
Before we start this one, we'll need to establish some important algebraic identities.
(B1) Rationalizing the Denominator
We aim to remove any square roots from the denominator. We multiply top and bottom of the fraction by the conjugate of the denominator. As a reminder, the conjugate of (3 + √2) is (3 − √2).
Example: We aim to rationalize the denominator of
\frac{5}{3+\sqrt{2}}
Multiplying top and bottom by (3 − √2) gives:
\frac{5}{3+\sqrt{2}}\times\frac{3-\sqrt{2}}{3-\sqrt{2}}=\frac{5(3-\sqrt{2})}{3^2-2}=\frac{5(3-\sqrt{2})}{7}
(B2) Rationalizing the Numerator
This is just the same as B1, but we are trying to remove square roots from the top of the fraction. We do that by multiplying top and bottom by the conjugate of the top.
(B3) Expanding the sine of a sum
This is one of the early trigonometry identities we learned:
sin (A + B) = sin A cos B + cos A sin B.
(B4) Limit of (cos θ - 1)/θ as x → 0
