prove that dy/dx of equation 'y = sinx' is cosx

Question

prove that dy/dx of equation 'y = sinx' is    cosx

Answer

Arre aisa hi hota hai Yr jb y = sinx hota hai then dy/dx = cosx hoti hai N jb y = cosx hota hai tabh dy/dx = - sinx hota hai

y= sinx
dy/dx = d(sinx) /dx
dy/dx = cosx ans.

Answer

Answer

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Answer

Derivation of dy/dx for y = sin(x)

To prove that the derivative of the equation y = sin(x) is cos(x), we will use the definition of the derivative and apply the properties of the sine function.

1. Definition of the derivative:
The derivative of a function f(x) with respect to x is defined as the limit of the difference quotient as the interval approaches zero:
dy/dx = lim(h→0) [(f(x+h) - f(x))/h]

2. Applying the definition to y = sin(x):
Let's substitute sin(x) for y in the definition of the derivative:
dy/dx = lim(h→0) [(sin(x+h) - sin(x))/h]

3. Applying the trigonometric identity:
Using the trigonometric identity for the difference of angles, we can rewrite the numerator as follows:
sin(x+h) - sin(x) = 2cos((x+h+x)/2)sin((x+h-x)/2)
= 2cos((2x+h)/2)sin(h/2)
= 2cos(x+(h/2))sin(h/2)

4. Simplifying the expression:
Now, let's substitute this expression back into the derivative formula:
dy/dx = lim(h→0) [2cos(x+(h/2))sin(h/2)/h]

5. Applying the limit:
To evaluate the limit as h approaches zero, we can use the trigonometric limit:
lim(h→0) sin(h)/h = 1

Since we have sin(h/2) in the numerator, we can rewrite the expression as follows:
dy/dx = lim(h→0) [2cos(x+(h/2))sin(h/2)/h] * (h/2)/(h/2)

6. Simplifying further:
dy/dx = lim(h→0) [2cos(x+(h/2))(sin(h/2)/(h/2))]

7. Applying the limit again:
Using the trigonometric limit once more, we have:
lim(h→0) sin(h/2)/(h/2) = 1

Therefore, the derivative simplifies to:
dy/dx = lim(h→0) [2cos(x+(h/2))]

8. Evaluating the limit:
Taking the limit as h approaches zero:
dy/dx = 2cos(x+(0/2))
= 2cos(x)

9. Final Result:
Hence, we have proved that the derivative of y = sin(x) is dy/dx = cos(x).

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