Ответы на вопрос:
f(x)=4x2+6x+3f′(x)=8x+6f′(x0)=f′(1)=8∗1+6=14f(x)=1+x2xf′(x)=(1+x2)21∗(1+x2)−x∗2x=(1+x2)21+x2−2x2=(1+x2)21−x2f′(0)=(1+02)21−02=11=1f(x)=(3x2+1)(3x2−1)=(3x2)2−12=9x4−1f′(x)=9∗4x3=36x3f′(1)=36∗13=36
\begin{gathered}f(x)=2x*cosx \\ f'(x)=2*cosx+2x*(-sinx)=2cosx-2xsinx \\ f'( \frac{ \pi }{4})=2cos\frac{ \pi }{4}-2*\frac{ \pi }{4}*sin\frac{ \pi }{4}=2*\frac{\sqrt{2}}{2}-2* \frac{ \pi }{4}*\frac{\sqrt{2}}{2}= \sqrt{2}(1- \frac{\pi}{4}) \end{gathered}f(x)=2x∗cosxf′(x)=2∗cosx+2x∗(−sinx)=2cosx−2xsinxf′(4π)=2cos4π−2∗4π∗sin4π=2∗22−2∗4π∗22=2(1−4π)
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