Calcular a derivada da função abaixo
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Vamos usar a regra da cadeia e a regra para derivada de funções exponenciais:
![f(x)=e^{\sqrt{x^2+4}} \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\sqrt{x^2+4}.ln(e)]' \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[(\sqrt{x^2+4})'.ln(e)+(\sqrt{x^2+4}).ln(e)'] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2}(x^2+4)^{-\frac{1}{2}}.(x^2+4)'.ln(e)+(\sqrt{x^2+4}).0] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2}(x^2+4)^{-\frac{1}{2}}.(2x+0).ln(e)+(\sqrt{x^2+4}).0] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2}(x^2+4)^{-\frac{1}{2}}.(2x).ln(e)] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2}.\frac{1}{\sqrt{x^2+4}}.(2x).ln(e)]](https://tex.z-dn.net/?f=f%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D+%5C%5C+%5C%5C+f%27%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%5B%5Csqrt%7Bx%5E2%2B4%7D.ln%28e%29%5D%27+%5C%5C+%5C%5C+f%27%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%5B%28%5Csqrt%7Bx%5E2%2B4%7D%29%27.ln%28e%29%2B%28%5Csqrt%7Bx%5E2%2B4%7D%29.ln%28e%29%27%5D+%5C%5C+%5C%5C+f%27%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%5B%5Cfrac%7B1%7D%7B2%7D%28x%5E2%2B4%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D.%28x%5E2%2B4%29%27.ln%28e%29%2B%28%5Csqrt%7Bx%5E2%2B4%7D%29.0%5D+%5C%5C+%5C%5C+f%27%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%5B%5Cfrac%7B1%7D%7B2%7D%28x%5E2%2B4%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D.%282x%2B0%29.ln%28e%29%2B%28%5Csqrt%7Bx%5E2%2B4%7D%29.0%5D+%5C%5C+%5C%5C+f%27%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%5B%5Cfrac%7B1%7D%7B2%7D%28x%5E2%2B4%29%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D.%282x%29.ln%28e%29%5D+%5C%5C+%5C%5C+f%27%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%5B%5Cfrac%7B1%7D%7B2%7D.%5Cfrac%7B1%7D%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%282x%29.ln%28e%29%5D "f(x)=e^{\sqrt{x^2+4}} \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\sqrt{x^2+4}.ln(e)]' \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[(\sqrt{x^2+4})'.ln(e)+(\sqrt{x^2+4}).ln(e)'] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2}(x^2+4)^{-\frac{1}{2}}.(x^2+4)'.ln(e)+(\sqrt{x^2+4}).0] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2}(x^2+4)^{-\frac{1}{2}}.(2x+0).ln(e)+(\sqrt{x^2+4}).0] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2}(x^2+4)^{-\frac{1}{2}}.(2x).ln(e)] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2}.\frac{1}{\sqrt{x^2+4}}.(2x).ln(e)]")
![f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2\sqrt{x^2+4}}.(2x).ln(e)] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.\frac{(2x).ln(e)}{2\sqrt{x^2+4}} \\ \\ f'(x)=\frac{e^{\sqrt{x^2+4}}.(2x).ln(e)}{2\sqrt{x^2+4}} \\ \\ f'(x)=\frac{e^{\sqrt{x^2+4}}.(2x).1}{2\sqrt{x^2+4}} \\ \\ f'(x)=\frac{e^{\sqrt{x^2+4}}.2x}{2\sqrt{x^2+4}} \\ \\ f'(x)=\frac{e^{\sqrt{x^2+4}}x}{\sqrt{x^2+4}}](https://tex.z-dn.net/?f=+f%27%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%5B%5Cfrac%7B1%7D%7B2%5Csqrt%7Bx%5E2%2B4%7D%7D.%282x%29.ln%28e%29%5D+%5C%5C+%5C%5C++f%27%28x%29%3De%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%5Cfrac%7B%282x%29.ln%28e%29%7D%7B2%5Csqrt%7Bx%5E2%2B4%7D%7D+%5C%5C+%5C%5C+f%27%28x%29%3D%5Cfrac%7Be%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%282x%29.ln%28e%29%7D%7B2%5Csqrt%7Bx%5E2%2B4%7D%7D+%5C%5C+%5C%5C+f%27%28x%29%3D%5Cfrac%7Be%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.%282x%29.1%7D%7B2%5Csqrt%7Bx%5E2%2B4%7D%7D+%5C%5C+%5C%5C+f%27%28x%29%3D%5Cfrac%7Be%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7D.2x%7D%7B2%5Csqrt%7Bx%5E2%2B4%7D%7D+%5C%5C+%5C%5C+f%27%28x%29%3D%5Cfrac%7Be%5E%7B%5Csqrt%7Bx%5E2%2B4%7D%7Dx%7D%7B%5Csqrt%7Bx%5E2%2B4%7D%7D "f'(x)=e^{\sqrt{x^2+4}}.[\frac{1}{2\sqrt{x^2+4}}.(2x).ln(e)] \\ \\ f'(x)=e^{\sqrt{x^2+4}}.\frac{(2x).ln(e)}{2\sqrt{x^2+4}} \\ \\ f'(x)=\frac{e^{\sqrt{x^2+4}}.(2x).ln(e)}{2\sqrt{x^2+4}} \\ \\ f'(x)=\frac{e^{\sqrt{x^2+4}}.(2x).1}{2\sqrt{x^2+4}} \\ \\ f'(x)=\frac{e^{\sqrt{x^2+4}}.2x}{2\sqrt{x^2+4}} \\ \\ f'(x)=\frac{e^{\sqrt{x^2+4}}x}{\sqrt{x^2+4}}")
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