Derivada por definición[tex]f(x) = \ sqrt{2x + 1} [ / tex]?
Derivada por definición [tex]f(x) = \ sqrt{2x + 1} [ / tex].
Derivada por definición [tex]f(x) = \ sqrt{2x + 1} [ / tex].
Pregunta de Matemáticas · 2 respuestas · 10 votos · mejor respuesta de Anyelizondo2871
En resumen
La derivada de f'(x) esta definida por : f'(x) = <img src="https://tex.z-dn.net/?f=%20%5Clim_%7Bh%20%5Cto%20%5C%200%7D%20%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20" /> reemplazando datos : <img src="https://tex.z-dn.net/?
La derivada de f'(x) esta definida por :
f'(x) = <img src="https://tex.z-dn.net/?f=%20%5Clim_%7Bh%20%5Cto%20%5C%200%7D%20%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20" />
reemplazando datos :
<img src="https://tex.z-dn.net/?f=%20%5Clim_%7Bh%20%5Cto%20%5C%200%7D%20%5Cfrac%7B%20%5Csqrt%7B2%28x%2Bh%29%2B1%7D%20-%20%5Csqrt%7B2x%2B1%7D%20%7D%7Bh%7D%20" />
multiplicando por el conjugado
<img src="https://tex.z-dn.net/?f=%20%5Clim_%7Bh%20%5Cto%20%5C%200%7D%20%5Cfrac%7B%20%5Csqrt%7B2x%2B2h%20%2B1%7D-%20%5Csqrt%7B2x%2B1%7D%20%7D%7Bh%7D%2A%20%5Cfrac%7B%28%20%5Csqrt%7B2x%2B2h%20%2B1%7D%20%2B%20%5Csqrt%7B2x%2B1%7D%20%29%7D%7B%28%20%5Csqrt%7B2x%2B2h%20%2B1%7D%2B%20%5Csqrt%7B2x%2B1%7D%20%29%7D%20" />
luego tenemos :
<img src="https://tex.z-dn.net/?f=%20%5Clim_%7Bh%20%5Cto%20%5C%200%7D%20%5Cfrac%7B2h%7D%7Bh%2A%28%20%5Csqrt%7B2x%2B2h%20%2B%201%7D%2B%20%5Csqrt%7B2x%2B1%7D%20%7D%20" />
eliminando h y reemplazando el limite tenemos :
<img src="https://tex.z-dn.net/?f=%20%5Cfrac%7B2%7D%7B%20%5Csqrt%7B2x%2B2%2A0%2B1%7D%2B%20%5Csqrt%7B2x%2B1%7D%20%20%7D%20" />
finalmente nos quedaria asi :
<img src="https://tex.z-dn.net/?f=f%27%28x%29%3D%20%20%5Cfrac%7B1%7D%7B%20%5Csqrt%7B2x%2B1%7D%20%7D%20" />.
<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%28x%29%3D%5Csqrt%7B2x%2B1%7D%20%5C%5C%5B2pt%5D%0A%5CRightarrow%20%5C%20f%28x%2Bh%29%3D%5Csqrt%7B2%28x%2Bh%29%2B1%7D%3D%5Csqrt%7B2x%2B2h%2B1%7D%20%20%5C%5C%5B8pt%5D%0Af%5E%7B%5Cprime%20%7D%28x%29%3D%5Clim_%7Bh%5Cto%200%7D%20%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh%7D%20%5C%5C%5B8pt%5D%0Af%5E%7B%5Cprime%20%7D%28x%29%3D%5Clim_%7Bh%5Cto%200%7D%20%5Cfrac%7B%5Csqrt%7B2x%2B2h%2B1%7D-%5Csqrt%7B2x%2B1%7D%7D%7Bh%7D%20%5C%5C%5B8pt%5D%0A%5Ctext%7BComo%20es%20un%20l%5C%27imite%20del%20tipo%200%2F0%20%20%2Cse%20multiplica%20en%20este%20caso%20por%20la%7D%5C%5C%0A%5Ctext%7Bconjugada%20del%20numerador.%7D%20%20%5C%5C%5B8pt%5D%0Af%5E%7B%5Cprime%20%7D%28x%29%3D%5Clim_%7Bh%5Cto%200%7D%5Cfrac%7B%5Csqrt%7B2x%2B2h%2B1%7D-%5Csqrt%7B2x%2B1%7D%7D%7Bh%7D%5Ccdot%20%5Cfrac%7B%5Csqrt%7B2x%2B2h%2B1%7D%2B%5Csqrt%7B2x%2B1%7D%7D%7B%5Csqrt%7B2x%2B2h%2B1%7D%2B%5Csqrt%7B2x%2B1%7D%7D%20%0A" />
<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Ctext%7BEn%20el%20numerador%20se%20aplica%20diferencia%20de%20cuadrados%20%7D%5C%5C%0A%28a%2Bb%29%28a-b%29%3Da%5E2-b%5E2%20%5C%5C%5B4pt%5D%0A%3D%5Clim_%7Bh%20%5Cto%200%7D%5C%2C%5Cfrac%7B%5Csqrt%7B2x%2B2h%2B1%7D%5E%7B%5C%2C2%7D-%5Csqrt%7B2x%2B1%7D%5E%7B%5C%2C2%7D%7D%7Bh%5Cleft%28%20%5Csqrt%7B2x%2B2h%2B1%7D%2B%5Csqrt%7B2x%2B1%7D%5C%2C%5Cright%29%7D%20%5C%5C%5B8pt%5D%0A%3D%5Clim_%7Bh%20%5Cto%200%7D%5C%2C%5Cfrac%7B%282x%2B2h%2B1%29-%282x%2B1%29%7D%7Bh%5Cleft%28%20%5Csqrt%7B2x%2B2h%2B1%7D%2B%5Csqrt%7B2x%2B1%7D%5C%2C%5Cright%29%7D%20%5C%5C%5B8pt%5D%0A%3D%5Clim_%7Bh%20%5Cto%200%7D%5C%2C%5Cfrac%7B2h%7D%7Bh%5Cleft%28%20%5Csqrt%7B2x%2B2h%2B1%7D%2B%5Csqrt%7B2x%2B1%7D%5C%2C%5Cright%29%7D%20%20%5C%5C%5B6pt%5D%0A%5Ctext%7BSe%20simplifica%20el%20factor%20%7Dh%20%20%0A%0A%20" />
<img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%3D%5Clim_%7Bh%5Cto%200%7D%5C%2C%20%5Cfrac%7B2%7D%7B%5Csqrt%7B2x%2B2h%2B1%7D%2B%5Csqrt%7B2x%2B1%7D%7D%20%5C%5C%5B8pt%5D%0A%3D%5Cfrac%7B2%7D%7B%5Csqrt%7B2x%2B%282%29%280%29%2B1%7D%2B%5Csqrt%7B2x%2B1%7D%7D%3D%5Cfrac%7B2%7D%7B2%5Csqrt%7B2x%2B1%7D%7D%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2x%2B1%7D%7D%5C%5C%5B8pt%5D%0A%5Ctext%7BPor%20lo%20tanto%20%3A%7D%20%5C%5C%5B8pt%5D%0Af%5E%7B%5Cprime%20%7D%28x%29%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2x%2B1%7D%7D" />.
La derivada de f'(x) esta definida por : f'(x) = <img src="https://tex.z-dn.net/?
Sí. Esta pregunta de Matemáticas tiene 2 respuestas de la comunidad; abajo puedes leerlas todas.