derive (sec^2 x . tang^2 x)
o resultado tem q ser: 2 sec^2 x .Tg x (2 tg^ 2 x + 1)
evelyn241294:
ⓝãⓞ sei
é ⓓⓞ ⓛⓘⓥⓡⓞ, ⓓⓔⓥⓔ ⓣⓔⓡ ⓐⓛⓖⓤⓜⓐ ⓒⓞⓘⓢⓐ ⓓⓘⓕⓔⓡⓔⓝⓣⓔ, ⓥⓐⓛⓔⓤ ⓟⓔⓛⓐ resposta
ⓣⓐ óⓣⓘⓜⓞ
:)
ⓣⓐ ⓒⓞⓜ ⓓⓔⓕⓔⓘⓣⓞ, ⓝ ⓒⓞⓝⓢⓘⓖⓞ ⓐⓡⓡⓤⓜⓐⓡ , ⓣⓞⓣⓔⓝⓣⓐⓝⓓⓞ ⓐⓠⓤⓘ, ⓕⓞⓘ ⓜⓐⓛ kkkk
Soluções para a tarefa
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Pra derivar vamos usar a regra da cadeia:
![y=sec^2(x).tg^2(x) \\ \\ y'=\frac{d}{dx} [sec^2(x)]. tg^2(x)+sec^2(x). \frac{d}{dx} [tg^2(x)] \\ \\ y'=2sec(x).tg(x).sec(x). tg^2(x)+sec^2(x). 2tg(x).sec^2(x) \\ \\ y'=2sec^2(x).tg(x). tg^2(x)+sec^2(x). 2tg(x).sec^2(x) \\ \\ y'=2sec^2(x). tg^3(x)+2sec^4(x).tg(x) \\ \\ y'=2sec^2(x).tg(x)[tg^2(x)+sec^2(x)]](https://tex.z-dn.net/?f=y%3Dsec%5E2%28x%29.tg%5E2%28x%29+%5C%5C+%5C%5C+y%27%3D%5Cfrac%7Bd%7D%7Bdx%7D+%5Bsec%5E2%28x%29%5D.+tg%5E2%28x%29%2Bsec%5E2%28x%29.+%5Cfrac%7Bd%7D%7Bdx%7D+%5Btg%5E2%28x%29%5D+%5C%5C+%5C%5C+y%27%3D2sec%28x%29.tg%28x%29.sec%28x%29.+tg%5E2%28x%29%2Bsec%5E2%28x%29.+2tg%28x%29.sec%5E2%28x%29+%5C%5C+%5C%5C+y%27%3D2sec%5E2%28x%29.tg%28x%29.+tg%5E2%28x%29%2Bsec%5E2%28x%29.+2tg%28x%29.sec%5E2%28x%29+%5C%5C+%5C%5C+y%27%3D2sec%5E2%28x%29.+tg%5E3%28x%29%2B2sec%5E4%28x%29.tg%28x%29+%5C%5C++%5C%5C++y%27%3D2sec%5E2%28x%29.tg%28x%29%5Btg%5E2%28x%29%2Bsec%5E2%28x%29%5D "y=sec^2(x).tg^2(x) \\ \\ y'=\frac{d}{dx} [sec^2(x)]. tg^2(x)+sec^2(x). \frac{d}{dx} [tg^2(x)] \\ \\ y'=2sec(x).tg(x).sec(x). tg^2(x)+sec^2(x). 2tg(x).sec^2(x) \\ \\ y'=2sec^2(x).tg(x). tg^2(x)+sec^2(x). 2tg(x).sec^2(x) \\ \\ y'=2sec^2(x). tg^3(x)+2sec^4(x).tg(x) \\ \\ y'=2sec^2(x).tg(x)[tg^2(x)+sec^2(x)]")
E chegamos a isso. :)
E chegamos a isso. :)
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