3.13-bo'lim : Logarifmik farqlash
5. \(f\left( x \right) = {\left( {2x - {{\bf{e}}^{8x}}} \right)^{\sin \left( {2x} \right)}}\) ning birinchi hosilasini toping.
Yechim
Biz faqat bir oz logarifmik farqlashimiz kerak, shuning uchun ikkala tomonning logarifmini oling va biroz soddalashtiring.
\[\ln \left[ {f\left( x \right)} \right] = \ln \left[ {{{\left( {2x - {{\bf{e}}^{8x}}} \right)}^{\sin \left( {2x} \right)}}} \right] = \sin \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right)\]2-qadam
\(x\) ga nisbatan har ikki tomonni farqlash uchun yashirin farqlashdan foydalaning. Mahsulotni o'ng tomonda boshqarishni unutmang.
\[\begin{align*}\frac{{f'\left( x \right)}}{{f\left( x \right)}} & = 2\cos \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right) + \sin \left( {2x} \right)\frac{{2 - 8{{\bf{e}}^{8x}}}}{{2x - {{\bf{e}}^{8x}}}}\\ & = 2\cos \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right) + \sin \left( {2x} \right)\frac{{2 - 8{{\bf{e}}^{8x}}}}{{2x - {{\bf{e}}^{8x}}}}\end{align*}\]3-qadam
Nihoyat, lotinni yeching va \(f\left( x \right)\) uchun tenglamani kiriting.
\[\begin{align*}f'\left( x \right) & = f\left( x \right)\left[ {2\cos \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right) + \sin \left( {2x} \right)\frac{{2 - 8{{\bf{e}}^{8x}}}}{{2x - {{\bf{e}}^{8x}}}}} \right]\\ & = \require{bbox} \bbox[2pt,border:1px solid black]{{{{\left( {2x - {{\bf{e}}^{8x}}} \right)}^{\sin \left( {2x} \right)}}\left[ {2\cos \left( {2x} \right)\ln \left( {2x - {{\bf{e}}^{8x}}} \right) + \sin \left( {2x} \right)\frac{{2 - 8{{\bf{e}}^{8x}}}}{{2x - {{\bf{e}}^{8x}}}}} \right]}}\end{align*}\]