Solve sinh(1)=

$$\sin h ( 1 ) =$$

Differentiate w.r.t. h

$$\cos(h)$$

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.

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

Use the Sum Formula for Sine.

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

Factor out \sin(h).

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

Rewrite the limit.

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

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

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

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

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

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

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

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

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

Use the Pythagorean Identity.

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

Rewrite the limit.

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

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

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

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

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

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

$$\cos(h)$$

Evaluate

$$\sin(h)$$