Odpowiedź:
[tex]\huge\boxed{f'(x)=\dfrac{2}{5}e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{x^4+3x^2-3x}{(x^2+1)^2}}[/tex]
Szczegółowe wyjaśnienie:
Skorzystamy ze wzorów:
[tex]\bigg(e^x\bigg)'=e^x\\\\\bigg[f\bigg(g(x)\bigg)\bigg]'=f'\bigg(g(x)\bigg)'\cdot g'(x)\\\\(ax^n)'=nax^{n-1}\\\\\bigg(\dfrac{f(x)}{g(x)}\bigg)'=\dfrac{f'(x)g(x)-f(x)g'(x)}{\left[g(x)\right]^2}[/tex]
[tex]\bigg(e^{\frac{2x^3+3}{5x^2+5}}\bigg)'=\bigg(e^{\frac{2x^3+3}{5x^2+5}}\bigg)'\cdot\left(\dfrac{2x^3+3}{5x^2+5}\right)'\\\\=e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{(2x^3+3)'(5x^2+5)-(2x^3+3)(5x^2+5)'}{(5x^2+5)^2}\\\\=e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{6x^2(5x^2+5)-(2x^3+3)\cdot10x}{(5x^2+5)^2}\\\\=e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{30x^4+30x^2-20x^4-30x}{(5x^2+5)^2}\\\\=e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{10x^4+30x^2-30x}{(5x^2+5)^2}\\\\=e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{10(x^4+3x^2-3x)}{[5(x^2+1)]^2}[/tex]
[tex]=e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{10(x^4+3x^2-3x)}{25(x^2+1)^2}\\\\=e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{2(x^4+3x^2-3x)}{5(x^2+1)^2}\\\\=\dfrac{2}{5}e^{\frac{2x^3+3}{5x^2+5}}\cdot\dfrac{x^4+3x^2-3x}{(x^2+1)^2}[/tex]
