A maior rede de estudos do Brasil

quem sabe essa, derivada de [sen(ln(tg(e^(x+2))))]^(x/2)

Cálculo IFACCAMP

9 resposta(s) - Contém resposta de Especialista

User badge image

RD Resoluções Verified user icon

Há mais de um mês

Para encontrarmos a derivada dessa função devemos realizar os calculos abaixo, utilizando a Regra da Cadeia:

\(\begin{align} & y={{\left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right]}^{x/2}} \\ & \ln y=\frac{x}{2}\ln \left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right] \\ & \frac{dy}{dx}=\frac{1}{2}{{\left\{ \sin (\ln (\tan ({{e}^{x+2}}))) \right\}}^{x/2}}\left( \ln (\tan ({{e}^{x+2}}))) \right)+ \\ & +\frac{x}{2}\frac{{{\left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right]}^{x/2}}\cos \left( \ln (\tan ({{e}^{x+2}})) \right){{\sec }^{2}}\left( {{e}^{x+2}} \right){{e}^{x+2}}}{\tan ({{e}^{x+2}})} \\ & \frac{dy}{dx}=\frac{1}{2}\ln {{\left\{ \sin (\ln (\tan ({{e}^{x+2}}))) \right\}}^{x/2}}+\frac{x}{2}\cdot \frac{1}{\left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right]}\left( \cos \left( \ln (\tan ({{e}^{x+2}})) \right) \right)\frac{1}{\tan ({{e}^{x+2}})}\left( {{\sec }^{2}}\left( {{e}^{x+2}} \right){{e}^{x+2}} \right) \\ & \frac{dy}{dx}=\frac{1}{2}\ln {{\left\{ \sin (\ln (\tan ({{e}^{x+2}}))) \right\}}^{x/2}} \\ & +\frac{x}{2}\cdot \frac{\left( \cos \left( \ln (\tan ({{e}^{x+2}})) \right) \right)\left( {{\sec }^{2}}\left( {{e}^{x+2}} \right){{e}^{x+2}} \right)}{\left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right]\cdot \left( \tan ({{e}^{x+2}}) \right)} \\ \end{align}\ \)

Portanto, o valor da derivada será:

\(\boxed{\frac{{dy}}{{dx}} = \frac{1}{2}\ln {{\left\{ {\sin \left( {\ln \left( {\tan \left( {{e^{x + 2}}} \right)} \right)} \right)} \right\}}^{x/2}} + \frac{x}{2} \cdot \frac{{\left( {\cos \left( {\ln \left( {\tan \left( {{e^{x + 2}}} \right)} \right)} \right)} \right)\left( {{{\sec }^2}\left( {{e^{x + 2}}} \right){e^{x + 2}}} \right)}}{{\left( {\sin \left( {\ln \left( {\tan \left( {{e^{x + 2}}} \right)} \right)} \right)} \right)\left( {\tan \left( {{e^{x + 2}}} \right)} \right)}}}\)

Para encontrarmos a derivada dessa função devemos realizar os calculos abaixo, utilizando a Regra da Cadeia:

\(\begin{align} & y={{\left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right]}^{x/2}} \\ & \ln y=\frac{x}{2}\ln \left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right] \\ & \frac{dy}{dx}=\frac{1}{2}{{\left\{ \sin (\ln (\tan ({{e}^{x+2}}))) \right\}}^{x/2}}\left( \ln (\tan ({{e}^{x+2}}))) \right)+ \\ & +\frac{x}{2}\frac{{{\left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right]}^{x/2}}\cos \left( \ln (\tan ({{e}^{x+2}})) \right){{\sec }^{2}}\left( {{e}^{x+2}} \right){{e}^{x+2}}}{\tan ({{e}^{x+2}})} \\ & \frac{dy}{dx}=\frac{1}{2}\ln {{\left\{ \sin (\ln (\tan ({{e}^{x+2}}))) \right\}}^{x/2}}+\frac{x}{2}\cdot \frac{1}{\left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right]}\left( \cos \left( \ln (\tan ({{e}^{x+2}})) \right) \right)\frac{1}{\tan ({{e}^{x+2}})}\left( {{\sec }^{2}}\left( {{e}^{x+2}} \right){{e}^{x+2}} \right) \\ & \frac{dy}{dx}=\frac{1}{2}\ln {{\left\{ \sin (\ln (\tan ({{e}^{x+2}}))) \right\}}^{x/2}} \\ & +\frac{x}{2}\cdot \frac{\left( \cos \left( \ln (\tan ({{e}^{x+2}})) \right) \right)\left( {{\sec }^{2}}\left( {{e}^{x+2}} \right){{e}^{x+2}} \right)}{\left[ \sin (\ln (\tan ({{e}^{x+2}}))) \right]\cdot \left( \tan ({{e}^{x+2}}) \right)} \\ \end{align}\ \)

Portanto, o valor da derivada será:

\(\boxed{\frac{{dy}}{{dx}} = \frac{1}{2}\ln {{\left\{ {\sin \left( {\ln \left( {\tan \left( {{e^{x + 2}}} \right)} \right)} \right)} \right\}}^{x/2}} + \frac{x}{2} \cdot \frac{{\left( {\cos \left( {\ln \left( {\tan \left( {{e^{x + 2}}} \right)} \right)} \right)} \right)\left( {{{\sec }^2}\left( {{e^{x + 2}}} \right){e^{x + 2}}} \right)}}{{\left( {\sin \left( {\ln \left( {\tan \left( {{e^{x + 2}}} \right)} \right)} \right)} \right)\left( {\tan \left( {{e^{x + 2}}} \right)} \right)}}}\)

Essa pergunta já foi respondida por um dos nossos especialistas