Giải câu 17 bài Ôn tập cuối năm.

\((u+v-w)'\)\(=u'+v'-w'\)
\((ku)'\)\(=ku'\)(k là hằng số)
\((uv)'\)\(=u'v+uv'\)
\(\left ( \frac{u}{v} \right )'\)\(=\frac{u'v-uv'}{v^2}(v=v(x)\neq 0)\)
\(\frac{1}{v}\)\(=-\frac{v'}{v^2}(v=v(x)\neq 0)\)
\(y'_x\)\(=y'_u.u'_x\)
\((x^n)’=nx^{n-1}\)\((u^n)’=nu^{n-1}.u’\)
\(\left ( \frac{1}{x} \right )’=-\frac{1}{x^2}\)\(\left ( \frac{1}{u} \right )’=-\frac{u’}{u^2}\)
\((\sqrt{x})’=\frac{1}{2\sqrt{x}}\)\((\sqrt{u})’=\frac{u’}{2\sqrt{u}}\)
\((sin\,x)’=cos\,x\)\((sin\,u)’=u’.cos\,u\)
\((cos\,x)’=-sin\,x\)\((cos\,u)’=-u’.sin\,u\)
\(\left ( tan\,x \right )'=\frac{1}{cos^2x}\)\(\left ( tan\,u \right )'=\frac{u’}{cos^2u}\)
\(\left ( cot\,x \right )'=-\frac{1}{sin^2x}\)\(\left ( cot\,u \right )'=-\frac{u’}{sin^2u}\)