• Matéria: Matemática
  • Autor: joelson81754
  • Perguntado 6 anos atrás

Calcule as derivadas da imagem

Anexos:

Respostas

respondido por: CyberKirito
1

\huge\boxed{\underline{\sf{Derivada~da~pot\hat{e}ncia}}}\\\sf{sendo~u~uma~func_\!\!,}\tilde{a}o~de~x}\\\sf{ent\tilde{a}o}\\\boxed{\boxed{\boxed{\boxed{\sf{\dfrac{d}{dx}~(u^n)=nu^{n-1}\cdot\dfrac{du}{dx}}}}}}}

\Large\boxed{\underline{\sf{Derivada~do~quociente~}}}\\\sf{sejam~u~e~v~duas~func_{\!\!,}\tilde{o}es~cont\acute{i}nuas}\\\sf{onde~a~2^{\underline{a}}~\acute{e}~n\tilde{a}o~nula}\\\sf{e~diferenci\acute{a}veis.}\\\sf{Ent\tilde{a}o}\\\boxed{\boxed{\boxed{\boxed{\sf{\left(\dfrac{u}{v}\right)'=\dfrac{u'\cdot v-u\cdot v'}{v^2}}}}}}

\boxed{\underline{\sf{Derivada~da~exponencial~de~base~qualquer}}}\\\sf{sendo~u~uma~func_{\!\!,}\tilde{a}o~de~x}\\\sf{Ent\tilde{a}o}\\\Large\boxed{\boxed{\boxed{\boxed{\sf{\dfrac{d}{dx}(a^u)=a^u\cdot~\ell n(a)\cdot\dfrac{du}{dx}}}}}}

\boxed{\underline{\sf{Derivada~da~func_{\!\!,}\tilde{a}o~logar\acute{i}tmica~de~base~qualquer}}}\\\sf{sendo~u~uma~func_{\!\!,}\tilde{a}o~de~x}\\\sf{Ent\tilde{a}o}\\\Large\boxed{\boxed{\boxed{\boxed{\sf{\dfrac{d}{dx}(log_au)=\dfrac{1}{u\cdot\ell n(a)}}\cdot\dfrac{du}{dx}}}}}

\dotfill

\boxed{\begin{array}{l}\tt{Calcule~as~seguintes~derivadas:}\end{array}}\\\tt{a)}~\sf{y=3x^5-11x^3-2}\\\Large\boxed{\boxed{\sf{\dfrac{dy}{dx}=15x^4-33x^2}}}\\\\\tt{d)}~\sf{y=\dfrac{(5x+1)log_3(x)}{2^x}}\\\sf{\dfrac{dy}{dx}=\dfrac{\left[5\cdot log_3(x)+(5x+1)\frac{1}{x\cdot\ell  n (3)}\right]\cdot2^x+(5x+1)\ell og_3x\cdot2^x\cdot\ell  n (2)}{\left(2^x\right)^2}}

\large\boxed{\sf{\dfrac{dy}{dx}=\dfrac{\left[5\cdot\ell og_3x+\frac{5x+1}{x\ell  n _3x}\right]\cdot2^x+(5x+1)\ell og_3x\cdot2^x\cdot\ell  n(2)}{2^{2x}}}}

\huge\boxed{\boxed{\boxed{\boxed{\boxed{\tt{\acute{O}timos~estudos~;)}}}}}}

Perguntas similares