How can we show that the derivative of sin x is cos x?  

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The derivative of a function f(x) is given by lim d--> 0 [(f(x + d) – f(x))/d]

For the function f(x) = sin x, f’(x) is given by

lim d--> 0 [(sin(x + d) – sin(x))/d]

We use the following relations here:

sin (a + b) = sin a*cos b + cos a* sin b

lim x--> 0 [sin x /x] = 1

lim x-->0[cos x / x] = 0 {we can derive these too, but that is not required here}

=> lim d--> 0 [(sin x* cos*d + cos x* sin d – sin(x))/d]

=> lim d--> 0 [(sin x* (cos*d – 1) + cos x* sin d)/d]

=> lim d--> 0 [(sin x* (cos*d – 1)]/d + lim d-->0 [cos x* sin d)/d]

apply the limit d-->0

=> sin x * 0 + cos x *1

=> cos x

Therefore for f(x) = sin x, f’(x) = cos x

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

To show that d/dx(sinx ) = cosx.

We know that d/dx {f(x) = f'(x) = Lt h->0 {f(x+h)-f(x)}/h.

Therefore (sinx)' = Lt h-> 0.  {sin(x+h)-sinx}/h .

(sinx)' = Lt {sinx*cosh+.*sinh - sinx}/h

(sinx)' = {(sinx)lt{(cosh) - 1}}/h + cosx*{Lt (sinh)/h}.

(sinx)' = sinx( 1-1} +cosx , as Lt h->0 {cosh -1 }/h = 0 and Lt h-> 0 (sinh)/h = 1.

(sinx)' = cosx.

Therefore (sinx)' = cosx .

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