Question

Fie f:(0,∞)→R, f(x)= $x^{1+sinx}$ . Derivata in orice x > 0, este ?

Answer 1

Răspuns:

Explicație pas cu pas:

aplicăm formula $(f^{g})'=(g*f^{g-1})*f'+(f^{g}*lnf)*g'=f^{g}*(f'*\frac{g}{f}+g'*lnf)~pentru~f>0,~unde~f=x,~g=1+sinx.\\\\f'(x)=(x^{1+sinx})'=x^{1+sinx}*(x'*\frac{1+sinx}{x}+(1+sinx)'*lnx)= x^{1+sinx}*(1*\frac{1+sinx}{x}+cosx*lnx)=x^{1+sinx}*(\frac{1+sinx}{x}+cosx*lnx).$

Answer 2

$f(x) = x^{1+\sin x} \\ \\\\\text{Logaritmez:}\\\\ \\ \ln \Big(f(x)\Big) = \ln \Big(x^{1+\sin x}\Big) \\ \\ \ln\Big(f(x)\Big) = (1+\sin x)\ln x\,\,\,\Big|'\\\\\\\text{Derivez:}\\\\\\ \dfrac{f'(x)}{f(x)} = \cos x\cdot \ln x + \dfrac{1+\sin x}{x}\\ \\ f'(x) = f(x)\cdot \Bigg[\cos x\cdot \ln x + \dfrac{1+\sin x}{x}\Bigg] \\ \\ \\ \Rightarrow f'(x) = x^{1+\sin x}\cdot \Bigg[\cos x\cdot \ln x+\dfrac{1+\sin x}{x}\Bigg]$