Solve sinx=

$$\sin x =$$

Differentiate w.r.t. x

$$\cos(x)$$

Steps Using Definition of a Derivative

For a function f\left(x\right), the derivative is the limit of \frac{f\left(x+h\right)-f\left(x\right)}{h} as h goes to 0, if that limit exists.

$$\frac{\mathrm{d}}{\mathrm{d}x}(\sin(x))=\left(\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}\right)$$

Use the Sum Formula for Sine.

$$\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}$$

Factor out \sin(x).

$$\lim_{h\to 0}\frac{\sin(x)\left(\cos(h)-1\right)+\cos(x)\sin(h)}{h}$$

Rewrite the limit.

$$\left(\lim_{h\to 0}\sin(x)\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\left(\lim_{h\to 0}\cos(x)\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$

Use the fact that x is a constant when computing limits as h goes to 0.

$$\sin(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(x)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$

The limit \lim_{x\to 0}\frac{\sin(x)}{x} is 1.

$$\sin(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(x)$$

To evaluate the limit \lim_{h\to 0}\frac{\cos(h)-1}{h}, first multiply the numerator and denominator by \cos(h)+1.

$$\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)=\left(\lim_{h\to 0}\frac{\left(\cos(h)-1\right)\left(\cos(h)+1\right)}{h\left(\cos(h)+1\right)}\right)$$

Multiply \cos(h)+1 times \cos(h)-1.

$$\lim_{h\to 0}\frac{\left(\cos(h)\right)^{2}-1}{h\left(\cos(h)+1\right)}$$

Use the Pythagorean Identity.

$$\lim_{h\to 0}-\frac{\left(\sin(h)\right)^{2}}{h\left(\cos(h)+1\right)}$$

Rewrite the limit.

$$\left(\lim_{h\to 0}-\frac{\sin(h)}{h}\right)\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$

The limit \lim_{x\to 0}\frac{\sin(x)}{x} is 1.

$$-\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$

Use the fact that \frac{\sin(h)}{\cos(h)+1} is continuous at 0.

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

Substitute the value 0 into the expression \sin(x)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(x).

$$\cos(x)$$

Evaluate

$$\sin(x)$$