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Câu hỏi: Tính \(f'\left( \pi \right)\) nếu \(f\left( x \right) = {{\sin x - x\cos x} \over {\cos x - x\sin x}}\)
Phương pháp giải
Tính đạo hàm sử dụng công thức \(\left( {\frac{u}{v}} \right)' = \dfrac{{u'v - uv'}}{{{v^2}}}\)
Lời giải chi tiết
+) Ta có:
$(\sin x-x \cdot \cos x)^{\prime}=(\sin x)^{\prime}-(x \cdot \cos x)^{\prime}$
$=\cos x-[1 \cdot \cos x+x \cdot(-\sin x)]=x \cdot \sin x$
$(\cos x-x \cdot \sin x)^{\prime}=(\cos x)^{\prime}-(x \cdot \sin x)^{\prime}$
$=-\sin x-(1 \cdot \sin x+x \cdot \cos x)=-2 \sin x-x \cdot \cos x$
$f^{\prime}(x)=\frac{(\sin x-x \cdot \cos x)^{\prime} \cdot(\cos x-x \cdot \sin x)-(\sin x-x \cdot \cos x) \cdot(\cos x-x \cdot \sin x)^{\prime}}{(\cos x-x \cdot \sin x)^{2}}$
$=\frac{x \cdot \sin x \cdot(\cos x-x \cdot \sin x)-(\sin x-x \cdot \cos x) \cdot(-2 \sin x-x \cdot \cos x)}{(\cos x-x \cdot \sin x)^{2}}$
$=\frac{x \cdot \sin x \cdot \cos x-x^{2} \cdot \sin ^{2} x+2 \sin ^{2} x+x \cdot \sin x \cdot \cos x-2 x \cdot \sin x \cdot \cos x-x^{2} \cdot \cos ^{2} x}{(\cos x-x \cdot \sin x)^{2}}$
$=\frac{-\left(x^{2} \sin ^{2} x+x^{2} \cos ^{2} x\right)+2 \sin ^{2} x}{(\cos x-x \cdot \sin x)^{2}}=\frac{-x^{2}+2 \sin ^{2} x}{(\cos x-x \cdot \sin x)^{2}}$
\(\Rightarrow f'\left( \pi \right) = \dfrac{{ - {\pi ^2} + 2{{\sin }^2}\pi }}{{{{\left({\cos \pi - \pi \sin \pi } \right)}^2}}} \) \(= \dfrac{{ - {\pi ^2} + 2.0}}{{{{\left( { - 1 - \pi. 0} \right)}^2}}} = - {\pi ^2}\)
Tính đạo hàm sử dụng công thức \(\left( {\frac{u}{v}} \right)' = \dfrac{{u'v - uv'}}{{{v^2}}}\)
Lời giải chi tiết
+) Ta có:
$(\sin x-x \cdot \cos x)^{\prime}=(\sin x)^{\prime}-(x \cdot \cos x)^{\prime}$
$=\cos x-[1 \cdot \cos x+x \cdot(-\sin x)]=x \cdot \sin x$
$(\cos x-x \cdot \sin x)^{\prime}=(\cos x)^{\prime}-(x \cdot \sin x)^{\prime}$
$=-\sin x-(1 \cdot \sin x+x \cdot \cos x)=-2 \sin x-x \cdot \cos x$
$f^{\prime}(x)=\frac{(\sin x-x \cdot \cos x)^{\prime} \cdot(\cos x-x \cdot \sin x)-(\sin x-x \cdot \cos x) \cdot(\cos x-x \cdot \sin x)^{\prime}}{(\cos x-x \cdot \sin x)^{2}}$
$=\frac{x \cdot \sin x \cdot(\cos x-x \cdot \sin x)-(\sin x-x \cdot \cos x) \cdot(-2 \sin x-x \cdot \cos x)}{(\cos x-x \cdot \sin x)^{2}}$
$=\frac{x \cdot \sin x \cdot \cos x-x^{2} \cdot \sin ^{2} x+2 \sin ^{2} x+x \cdot \sin x \cdot \cos x-2 x \cdot \sin x \cdot \cos x-x^{2} \cdot \cos ^{2} x}{(\cos x-x \cdot \sin x)^{2}}$
$=\frac{-\left(x^{2} \sin ^{2} x+x^{2} \cos ^{2} x\right)+2 \sin ^{2} x}{(\cos x-x \cdot \sin x)^{2}}=\frac{-x^{2}+2 \sin ^{2} x}{(\cos x-x \cdot \sin x)^{2}}$
\(\Rightarrow f'\left( \pi \right) = \dfrac{{ - {\pi ^2} + 2{{\sin }^2}\pi }}{{{{\left({\cos \pi - \pi \sin \pi } \right)}^2}}} \) \(= \dfrac{{ - {\pi ^2} + 2.0}}{{{{\left( { - 1 - \pi. 0} \right)}^2}}} = - {\pi ^2}\)