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Correction exercice
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Correction exercice I

Calculons les dérivées des fonctions suivantes

1. \(y = \arcsin \frac{x}{a}\)
Nous partons du fait que \(\left( {\arcsin x} \right)' = \) \(\frac{1}{{\sqrt {1 + {x^2}} }}\) et de la définition de la dérivée de la fonction réciproque
Posons \(u(x) = \frac{x}{a}\) ainsi, \(y = \arcsin \left( {u(x)} \right)\)
\(y' = \left( {\arcsin \left( {u(x)} \right)} \right)'\) \( = \frac{1}{{\sqrt {1 - {u^2}} }}u'(x) = \) \(\frac{1}{{\sqrt {1 - {{\left( {\frac{x}{a}} \right)}^2}} }}\frac{1}{a} = \) \(\frac{1}{{\sqrt {{a^2} - {x^2}} }}\)

\(y' = \frac{1}{{\sqrt {{a^2} - {x^2}} }}\)

2. \(y = {\left( {\arcsin x} \right)^2}\)
\(y' = \) \(2\arcsin x\left( {\arcsin x} \right)'\) \( = 2\arcsin x.\) \(\frac{1}{{\sqrt {1 - {x^2}} }}\)

\(y' = \frac{{2\arcsin x}}{{\sqrt {1 - {x^2}} }}\)

3. \(y = \) \(\arctan \left( {{x^2} + 1} \right)\)
\(u(x) = {x^2} + 1\)
\(u'(x) = 2x\)
\(y' = \frac{1}{{1 + {u^2}}}\) \(u'(x) = \) \(\frac{1}{{1 + {{\left( {{x^2} + 1} \right)}^2}}}\) \(2x\)

\(y' = \frac{{2x}}{{1 + {{\left( {{x^2} + 1} \right)}^2}}}\)

4. \(y = \arcsin \frac{{2x}}{{1 - {x^2}}}\)
\(u(x) = \frac{{2x}}{{1 - {x^2}}}\)
\(u'(x) = \) \(\frac{{2\left( {1 - {x^2}} \right) + 4{x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}\) \( = \frac{{2 + 2{x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}\)
\(y' = \frac{1}{{\sqrt {1 - {u^2}} }}\) \(u'(x) = \) \(\frac{1}{{\sqrt {1 - {{\left( {\frac{{2x}}{{1 - {x^2}}}} \right)}^2}} }}\) \(\frac{{2 + 2{x^2}}}{{{{\left( {1 - {x^2}} \right)}^2}}}\)

\(y' = \) \(\frac{{2 + 2{x^2}}}{{\left( {1 - {x^2}} \right)\sqrt {{x^4} - 6{x^2} + 1} }}\)

5. \(y = \arccos {x^2}\)
\(u(x) = {x^2}\)
\(u'(x) = 2x\)
\(y' = \frac{1}{{\sqrt {1 - {u^2}} }}\) \(u'(x) = - \) \(\frac{1}{{\sqrt {1 - {{\left( {{x^2}} \right)}^2}} }}2x\)

\(y' = - \frac{{2x}}{{\sqrt {1 - {x^4}} }}\)

6. \(y = \frac{1}{{\sqrt 3 }}\arccos \frac{{x\sqrt 3 }}{{1 - {x^2}}}\)
\(u(x) = \frac{{x\sqrt 3 }}{{1 - {x^2}}}\)
\(u'(x) = \) \(\frac{{\sqrt 3 \left( {1 + {x^2}} \right)}}{{{{\left( {1 - {x^2}} \right)}^2}}}\)
\(y' = \frac{1}{{1 + {u^2}}}u'(x)\)

\(y' = \) \(\frac{{1 + {x^2}}}{{{x^4} + {x^2} + 1}}\)

7. \(y = \arcsin \sqrt {\sin x} \)
\(u(x) = \sqrt {\sin x} \)
\(u'(x) = \) \(\frac{{\cos x}}{{2\sqrt {\sin x} }}\)
\(y' = \) \(\frac{1}{{\sqrt {1 - {u^2}} }}u'(x)\)

\(y' = \) \(\frac{{\cos x}}{{2\sqrt {\sin x} \sqrt {1 - \sin x} }}\)

8. \(y = \) \(\arctan \frac{{4\sin x}}{{3 + 5\cos x}}\)

\(u(x) = \) \(\frac{{4\sin x}}{{3 + 5\cos x}}\)
\(u'(x) = \) \(\frac{{4\cos x(3 + 5\cos x)}}{{{{(3 + 5\cos x)}^2}}}\) \( + \frac{{5{{\sin }^2}x}}{{{{(3 + 5\cos x)}^2}}}\)
\(y' = \frac{1}{{1 + {u^2}}}u'(x)\)

\(y' = \) \(\frac{{4\cos x(3 + 5\cos x)}}{{16{{\sin }^2}x + {{(3 + 5\cos x)}^2}}}\) \( + \) \(\frac{{5{{\sin }^2}x}}{{16{{\sin }^2}x + {{(3 + 5\cos x)}^2}}}\)

Exercice II

1. Calcule de \({S_n}\)
En réorganisant \(\tan {u_n}\), nous avons : \(\tan {u_n} = \) \(\frac{{n + 1 - n}}{{1 + n(n + 1)}}\)
En posant : \(\tan {\alpha _n} = n + 1\) avec \(0 \prec {\alpha _n} \prec \frac{\pi }{2}\) et \(\tan {\beta _n} = n\) avec \(0 \prec {\beta _n} \prec \frac{\pi }{2}\) , nous avons
\(\tan {u_n} = \) \(\frac{{\tan {\alpha _n} - \tan {\beta _n}}}{{1 + \tan {\beta _n}.\tan {\alpha _n}}}\) \( = \tan \left( {{\alpha _n} - {\beta _n}} \right)\)
D’où \({{u_n} = {\alpha _n} - {\beta _n}}\)
• \({u_n} = {\alpha _n} - {\beta _n}\) \( = \arctan \left( {n + 1} \right)\) \( - \arctan \left( n \right)\)
• \({u_{n - 1}} = {\alpha _{n - 1}} - {\beta _{n - 1}}\) \( = \arctan \left( n \right) - \) \(\arctan \left( {n - 1} \right)\)
• \({u_{n - 2}} = {\alpha _{n - 2}} - {\beta _{n - 2}}\) \( = \arctan \left( {n-1} \right) - \) \(\arctan \left( {n - 2} \right)\)
• ….
• \({u_1} = {\alpha _1} - {\beta _1}\) \( = \arctan \left( 2 \right) - \) \(\arctan \left( 1 \right)\)
Par addition membre à membre de ces n égalités, nous obtenons
\({S_n} = \arctan \left( {n + 1} \right)\) \( - \arctan \left( 1 \right) = \) \(\arctan \left( {n + 1} \right)\) \( - \frac{\pi }{4}\)
2. Calcule de la limite de \({S_n}\) lorsque \(n \to \infty \)
\(\mathop {\lim }\limits_{n \to \infty } {S_n} = \) \(\mathop {\lim }\limits_{n \to \infty } \left( {\arctan \left( {n + 1} \right)} \right)\) \( - \arctan \left( 1 \right) = \) \(\frac{\pi }{2} - \frac{\pi }{4} = \frac{\pi }{4}\).