derivar e simplificar
senx.cosx/x³
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Oi Elaine :) . Segue a derivada da função
![y= \frac{sen(x)cox(x)}{x^3} \\ \\ y= \frac{ \frac{d}{dx} (sen(x)cos(x)).x^3-sen(x) cos(x)\frac{d}{dx} (x^3)}{(x^3)^2} \\ \\ y'= \frac{ \frac{d}{dx} [\frac{d}{dx}(sen(x)).cos(x)+sen(x).\frac{d}{dx}(cos(x))].x^3-sen(x)cos(x)3x^2}{x^6} \\ \\ y'= \frac{ x^3 [cos(x).cos(x)+sen(x).-sen(x)]-sen(x)cos(x)3x^2}{x^6} \\ \\ y'= \frac{ x [cos^2(x)-sen^2(x)]-sen(x)cos(x)3x^2}{x^6} \\ \\ y'= \frac{ x^3 [cos^2(x)-sen^2(x)]-3sen(x)cos(x)}{x^4}](https://tex.z-dn.net/?f=y%3D+%5Cfrac%7Bsen%28x%29cox%28x%29%7D%7Bx%5E3%7D++%5C%5C++%5C%5C+y%3D+%5Cfrac%7B+%5Cfrac%7Bd%7D%7Bdx%7D+%28sen%28x%29cos%28x%29%29.x%5E3-sen%28x%29+cos%28x%29%5Cfrac%7Bd%7D%7Bdx%7D+%28x%5E3%29%7D%7B%28x%5E3%29%5E2%7D+%5C%5C++%5C%5C++y%27%3D+%5Cfrac%7B+%5Cfrac%7Bd%7D%7Bdx%7D+%5B%5Cfrac%7Bd%7D%7Bdx%7D%28sen%28x%29%29.cos%28x%29%2Bsen%28x%29.%5Cfrac%7Bd%7D%7Bdx%7D%28cos%28x%29%29%5D.x%5E3-sen%28x%29cos%28x%293x%5E2%7D%7Bx%5E6%7D+%5C%5C++%5C%5C++y%27%3D+%5Cfrac%7B+x%5E3+%5Bcos%28x%29.cos%28x%29%2Bsen%28x%29.-sen%28x%29%5D-sen%28x%29cos%28x%293x%5E2%7D%7Bx%5E6%7D+%5C%5C++%5C%5C++y%27%3D+%5Cfrac%7B+x+%5Bcos%5E2%28x%29-sen%5E2%28x%29%5D-sen%28x%29cos%28x%293x%5E2%7D%7Bx%5E6%7D+%5C%5C++%5C%5C+y%27%3D+%5Cfrac%7B+x%5E3+%5Bcos%5E2%28x%29-sen%5E2%28x%29%5D-3sen%28x%29cos%28x%29%7D%7Bx%5E4%7D "y= \frac{sen(x)cox(x)}{x^3} \\ \\ y= \frac{ \frac{d}{dx} (sen(x)cos(x)).x^3-sen(x) cos(x)\frac{d}{dx} (x^3)}{(x^3)^2} \\ \\ y'= \frac{ \frac{d}{dx} [\frac{d}{dx}(sen(x)).cos(x)+sen(x).\frac{d}{dx}(cos(x))].x^3-sen(x)cos(x)3x^2}{x^6} \\ \\ y'= \frac{ x^3 [cos(x).cos(x)+sen(x).-sen(x)]-sen(x)cos(x)3x^2}{x^6} \\ \\ y'= \frac{ x [cos^2(x)-sen^2(x)]-sen(x)cos(x)3x^2}{x^6} \\ \\ y'= \frac{ x^3 [cos^2(x)-sen^2(x)]-3sen(x)cos(x)}{x^4}")
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