How to solve cos beta X

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

$$cos\beta_{X}$$

Differentiate w.r.t. β_X

$-\sin(\beta _{X})$

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.

$$\frac{\mathrm{d}}{\mathrm{d}\beta _{X}}(\cos(\beta _{X}))=\left(\lim_{h\to 0}\frac{\cos(\beta _{X}+h)-\cos(\beta _{X})}{h}\right)$$

Use the Sum Formula for Cosine.

$$\lim_{h\to 0}\frac{\cos(h+\beta _{X})-\cos(\beta _{X})}{h}$$

Factor out $\cos(\beta _{X})$.

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

Rewrite the limit.

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

Use the fact that $\beta _{X}$ is a constant when computing limits as $h$ goes to $0$.

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

The limit $\lim_{\beta _{X}\to 0}\frac{\sin(\beta _{X})}{\beta _{X}}$ is $1$.

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

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 $\lim_{\beta _{X}\to 0}\frac{\sin(\beta _{X})}{\beta _{X}}$ 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 $\cos(\beta _{X})\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)-\sin(\beta _{X})$.

$$-\sin(\beta _{X})$$

Show Solution

Evaluate

$\cos(\beta _{X})$