How to solve f(x)= sinx

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

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Question

$$f(x)= \sin x$$

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

$$\cos(x)$$

Show Solution

Evaluate

$\sin(x)$