Question

Se n for um inteiro positivo, demonstre que $\frac{d}{dx} (sen^n x cos nx ) = n sen^n-1 x con(n+1)x$

Answer 1

$\boxed{\boxed{f(x)=sen^n(x)cos(nx)}}$

regra do produto
$\boxed{(U*V)'=U'*V+U*V'}$

temos:

$\bmatrix U=sen^n(x)\\\\U'=n*sen^{n-1}(x)*cos(x)\\\\V=cos(n*x)\\\\V'=-sen(n*x)*n\end$

então:
$f'(x)=n*sen^{n-1}(x)*cos(x)*cos(n*x) + sen^n(x)*-sen(n*x)*n\\\\ f'(x)= nsen^{n-1}(x)cos(x)cos(nx)-nsen^n(x)sen(nx)\\\\\boxed{\boxed{f'(x)=nsen^{n-1}[cos(x)cos(nx)-sen(x)sen(nx)]}}$


cos(a+b)=cos(a)cos(b)-sen(a)sen(b)


$f'(x)=nsen^{n-1}[cos(x+nx)]}\\\\\boxed{\boxed{f'(x)=nsen^{n-1}cos(\;(1+n)x\;)}}$