Oblicz pochodne podanych funkcji

[anka]

Oblicz pochodne podanych funkcji:
a)
\(f(x) =\frac{e^{2x} cos^ {2}x+4x^3}{sin 4x+2 tg x}\)

b)
\(f(x) = arccos \sqrt{\frac{ 1-x^ 2}{1+x^2}}\)

c)
\(f(x) = (sin x)^{x ^2 -1}\)

d)
\(f(x) = arctg(sin(1 + e^ {3x}))\)

Za podpowiedzi dziękuję, potrzebne mi są tzw gotowce

[patryk00714]

d)

\(f'(x)=\frac{1}{1+(sin(1+e^{3x}))^2} \cdot (sin(1+e^{3x}))'=\frac{1}{1+(sin(1+e^{3x}))^2} \cdot cos(1+e^{3x}) \cdot (1+e^{3x})'=\\=\frac{cos(1+e^{3x})}{1+(sin(1+e^{3x}))^2} \cdot 3e^{3x}=\frac{3e^{3x}cos(1+e^{3x})}{1+(sin(1+e^{3x}))^2}\)

[patryk00714]

\(c) f(x)=(sinx)^{x^2-1}=e^{(x^2-1)ln(sinx)}\)

stąd

\(f'(x)=e^{(x^2-1)ln(sinx)} \cdot [(x^2-1)ln(sinx)]'=(sinx)^{x^2-1} \cdot [ 2xln(sinx)+(x^2-1) \cdot \frac{cosx}{sinx}]= \\ =(sinx)^{x^2-1} \cdot [ 2xln(sinx)+(x^2-1) \cdot ctgx]\)

[patryk00714]

b) \(f'(x)=-\frac{1}{\sqrt{1-\frac{1-x^2}{1+x^2}}} \cdot \left( \sqrt{\frac{1-x^2}{1+x^2} \right)'=\frac{1}{\sqrt{\frac{2x^2}{1+x^2}}} \cdot \frac{1}{2\sqrt{\frac{1-x^2}{1+x^2}}} \cdot \left(\frac{1-x^2}{1+x^2} \right)' = \\ = \frac{1+x^2}{2\sqrt{2}|x|\sqrt{1-x^2}} \cdot \frac{-2x(1+x^2)-(1-x^2)2x}{(1+x^2)^2}=\frac{1+x^2}{2\sqrt{2}|x|\sqrt{1-x^2}} \cdot \frac{4x}{(1+x^2)^2}= \frac{2x}{\sqrt{2}|x|\sqrt{1-x^2}(1+x^2)} =\\= \frac{\sqrt{2}x}{|x|\sqrt{1-x^2}(1+x^2)}\)

[patryk00714]

a) \(f(x)=\frac{e^{2x}cos^2x+4x^3}{sin4x+2tgx}\)

\(f'(x)=\frac{ \left(e^{2x}cos^2x+4x^3 \right)' \cdot \left(sin4x+2tgx \right)- \left(e^{2x}cos^2x+4x^3 \right) \cdot \left(sin4x+2tgx \right)'}{ \left(sin4x+2tgx \right)^2}=\)

\(= \frac{ \left[\left(e^{2x}cos^2x \right)'+12x^2 \right] \cdot \left(sin4x+2tgx \right)-\left(e^{2x}cos^2x+4x^3 \right) \cdot \left(4cos4x+\frac{2}{cos^2x} \right) }{\left(sin4x+2tgx \right)^2}=\)

\(= \frac{ \left(2e^{2x}cos^2x+e^{2x} \cdot 2cosx(-sinx)+12x^2 \right)- \left(e^{2x}cos^2x+4x^3 \right) \cdot \left(4cos4x+\frac{2}{cos^2x} \right) }{\left(sin4x+2tgx \right)^2}=\)

\(= \frac{ \left[e^{2x} \left(2cos^2x-sin2x \right)+12x^2 \right] -\left(e^{2x}cos^2x+4x^3 \right) \cdot \left(4cos4x+\frac{2}{cos^2x} \right) }{\left(sin4x+2tgx \right)^2}\)