Question

Determine a derivada da função f(×)=1-✓1+cos²(e×)

Answer 1

Resposta:

$f(x)=1-\sqrt{1+cos^2e^x} \\\\f'(x)=0-\frac{0+2cose^x.(-sene^x).e^x}{2\sqrt{1+cos^2e^x} } \\\\f'(x)=\frac{e^xsen2e^x}{\sqrt{1+cos^2e^x} }$

Answer 2

Resposta:

$\sf f(x)=1-\sqrt{1+cos^2e^x}$

$\sf\dfrac{df}{dx}=\dfrac{d}{dx}(1-\sqrt{1+cos^2e^x})$

$\sf\dfrac{df}{dx}=\dfrac{d}{dx}(1)-\dfrac{d}{dx}(\sqrt{1+cos^2e^x})$

$\sf\dfrac{df}{dx}=0-\dfrac{d}{dx}(\sqrt{1+cos^2e^x})$

$\sf\dfrac{df}{dx}=-\dfrac{d}{dx}(\sqrt{1+cos^2e^x})$

Pela regra da cadeia, na qual

$\boxed{\sf \dfrac{d}{dx}(f(u))=\dfrac{d}{du}(f(u))\cdot \dfrac{d}{dx}(u)}$

, faça u = 1 + cos²eˣ:

$\sf\dfrac{df}{dx}=-\bigg[\dfrac{d}{du}(\sqrt{u})\cdot\dfrac{d}{dx}(u)\bigg]$

$\sf\dfrac{df}{dx}=-\bigg[\dfrac{d}{du}(u^{\frac{1}{2}})\cdot\dfrac{d}{dx}(u)\bigg]$

$\sf\dfrac{df}{dx}=-\bigg[\frac{1}{2}\cdot (1+cos^2e^x)^{\frac{1}{2}-1}\cdot\dfrac{d}{dx}(1+cos^2e^x)\bigg]$

$\sf\dfrac{df}{dx}=-\bigg[\dfrac{1}{2}(1+cos^2e^x)^{-\frac{1}{2}}\cdot\bigg(\dfrac{d}{dx}(1)+\dfrac{d}{dx}(cos^2e^x)\bigg)\bigg]$

$\sf\dfrac{df}{dx}=-\bigg[\dfrac{1}{2}\cdot\dfrac{1}{(1+cos^2e^x)^{\frac{1}{2}}}\cdot\bigg(0+\dfrac{d}{dx}(cos^2e^x)\bigg)\bigg]$

$\sf\dfrac{df}{dx}=-\dfrac{1}{2\sqrt{1+cos^2e^x}}\cdot\dfrac{d}{dx}(cos^2e^x)$

Aplicando a regra da cadeia novamente, faça v = cos eˣ:

$\sf\dfrac{df}{dx}=-\dfrac{1}{2\sqrt{1+cos^2e^x}}\cdot\dfrac{d}{dv}(v^2)\cdot\dfrac{d}{dx}(v)$

$\sf\dfrac{df}{dx}=-\dfrac{1}{2\sqrt{1+cos^2e^x}}\cdot2v\cdot\dfrac{d}{dx}(v)$

$\sf\dfrac{df}{dx}=-\dfrac{1}{2\sqrt{1+cos^2e^x}}\cdot2cos\,e^x\cdot\dfrac{d}{dx}(cos\,e^x)$

Novamente, agora faça w = eˣ:

$\sf\dfrac{df}{dx}=-\dfrac{1}{2\sqrt{1+cos^2e^x}}\cdot2cos\,e^x\cdot\dfrac{d}{dw}(cos\,w)\cdot\dfrac{d}{dx}(w)$

$\sf\dfrac{df}{dx}=-\dfrac{1}{2\sqrt{1+cos^2e^x}}\cdot2cos\,e^x\cdot(-\,sen\,w)\cdot\dfrac{d}{dx}(w)$

$\sf\dfrac{df}{dx}=\dfrac{1}{2\sqrt{1+cos^2e^x}}\cdot2cos\,e^x\cdot sen\,e^x\cdot\dfrac{d}{dx}(e^x)$

$\sf\dfrac{df}{dx}=\dfrac{1}{2\sqrt{1+cos^2e^x}}\cdot2cos\,e^x\cdot sen\,e^x\cdot e^x$

$\sf\dfrac{df}{dx}=\dfrac{2cos\,e^x\cdot sen\,e^x\cdot e^x}{2\sqrt{1+cos^2e^x}}$

$\red{\boxed{\sf\dfrac{df}{dx}=\dfrac{cos\,e^x\cdot sen\,e^x\cdot e^x}{\sqrt{1+cos^2e^x}}}$