Ответ:
Логарифмическое дифференцирование .
[tex]\bf 1)\ \ y=x^{sinx^2}\\\\lny=ln\Big(x^{sinx^2}\Big)\\\\lny=sinx^2\cdot lnx\\\\\dfrac{y'}{y}=cosx^2\cdot 2x\cdot lnx+sinx^2\cdot \dfrac{1}{x}\\\\y'=y\cdot \Big(cosx^2\cdot 2x\cdot lnx+sinx^2\cdot \dfrac{1}{x}\Big)\\\\y'=x^{sinx^2}\cdot \Big(cosx^2\cdot 2x\cdot lnx+sinx^2\cdot \dfrac{1}{x}\Big)[/tex]
[tex]\bf 2)\ \ y=(4x)^{x^4+1}\\\\lny=ln\Big((4x)^{x^4+1}\Big)\\\\lny=(x^4+1)\cdot ln(4x)\\\\\dfrac{y'}{y}=4x^3\cdot ln(4x)+(x^4+1)\cdot \dfrac{4}{4x}\\\\y'=y\cdot \Big(4x^3\cdot ln(4x)+(x^4+1)\cdot \dfrac{1}{x}\Big)\\\\y'=(4x)^{x^4+1}\cdot \Big(4x^3\cdot ln(4x)+(x^4+1)\cdot \dfrac{1}{x}\Big)[/tex]
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Answers & Comments
Ответ:
Логарифмическое дифференцирование .
[tex]\bf 1)\ \ y=x^{sinx^2}\\\\lny=ln\Big(x^{sinx^2}\Big)\\\\lny=sinx^2\cdot lnx\\\\\dfrac{y'}{y}=cosx^2\cdot 2x\cdot lnx+sinx^2\cdot \dfrac{1}{x}\\\\y'=y\cdot \Big(cosx^2\cdot 2x\cdot lnx+sinx^2\cdot \dfrac{1}{x}\Big)\\\\y'=x^{sinx^2}\cdot \Big(cosx^2\cdot 2x\cdot lnx+sinx^2\cdot \dfrac{1}{x}\Big)[/tex]
[tex]\bf 2)\ \ y=(4x)^{x^4+1}\\\\lny=ln\Big((4x)^{x^4+1}\Big)\\\\lny=(x^4+1)\cdot ln(4x)\\\\\dfrac{y'}{y}=4x^3\cdot ln(4x)+(x^4+1)\cdot \dfrac{4}{4x}\\\\y'=y\cdot \Big(4x^3\cdot ln(4x)+(x^4+1)\cdot \dfrac{1}{x}\Big)\\\\y'=(4x)^{x^4+1}\cdot \Big(4x^3\cdot ln(4x)+(x^4+1)\cdot \dfrac{1}{x}\Big)[/tex]
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