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find derivative of sin x by first principle.

Posted by Jayesh Aggarwal 8 years, 11 months ago

CBSE > Class 11 > Mathematics

1 answers

Naveen Sharma 8 years, 11 months ago

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:

$\lim_{h\hspace{1}\to\hspace{1}0}\frac{f(x+h)-f(x)}{h}$

$=\lim_{h\hspace{1}\to\hspace{1}0}\large{\frac{2\sin\left(\frac{h}{2}\right)\cos\left(x+\frac{h}{2}\right)}{h}}$

We make use of (3), fraction on a fraction, to bring that 2 out front down to the bottom:

$=\lim_{h\hspace{1}\to\hspace{1}0}\large{\frac{\sin\left(\frac{h}{2}\right)\cos\left(x+\frac{h}{2}\right)}{\frac{h}{2}}}$

Now, the limit of a product is the product of the limits, so we can write this as:

$=\lim_{h\hspace{1}\to\hspace{1}0}\frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}$ $\times{\lim_{h\hspace{1}\to\hspace{1}0}\cos\left(x+\frac{h}{2}\right)}$

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

$\frac{d}{dx}\sin{x}=\cos{x}$

0Thank You

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