第 12 类 RD Sharma 解决方案 – 第 11 章微分 – 练习 11.5 |设置 1

问题 1. 对 x 求微分 y = x ^1/ x。

解决方案：

We have, 

=> y = x^1/x

On taking log of both the sides, we get,

=> log y = log x^1/x

=> log y = (1/x) (log x)

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(\frac{logx}{x})$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}(\frac{1}{x})+logx(\frac{-1}{x^2})$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{x^2}-\frac{logx}{x^2}$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{1-logx}{x^2}$

=> $\frac{dy}{dx}=\frac{(1-logx)y}{x^2}$

=> $\frac{dy}{dx}=\frac{(1-logx)x^{\frac{1}{x}}}{x^2}$

问题 2. 对 x 微分 y = x ^sin x。

解决方案：

We have, 

=> y = x^{sin x}

On taking log of both the sides, we get,

=> log y = log x^{sin x}

=> log y = sin x log x 

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(sinxlogx)$

=> $\frac{1}{y}\frac{dy}{dx}=sinx(\frac{1}{x})+logx(cosx)$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{sinx}{x}+logxcosx$

=> $\frac{dy}{dx}=y\left[\frac{sinx}{x}+logxcosx\right]$

=> $\frac{dy}{dx}=x^{sinx}\left[\frac{sinx}{x}+logxcosx\right]$

问题 3. 对 x 求微分 y = (1 + cos x) ^x 。

解决方案：

We have, 

=> y = (1 + cos x)^x

On taking log of both the sides, we get,

=> log y = log (1 + cos x)^x 

=> log y = x log (1 + cos x)

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (1 + cos x)]$

=> $\frac{1}{y}\frac{dy}{dx}=-xsinx(\frac{1}{1+cosx})+log(1+cosx)(1)$

=> $\frac{1}{y}\frac{dy}{dx}=-xsinx(\frac{1}{1+cosx})+log(1+cosx)(1)$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{-xsinx}{1+cosx}+log(1+cosx)$

=> $\frac{dy}{dx}=y\left[log(1+cosx)-\frac{xsinx}{1+cosx}\right]$

=> $\frac{dy}{dx}=(1+cos x)^x\left[log(1+cosx)-\frac{xsinx}{1+cosx}\right]$

问题 4. 区分$y=x^{cos^{-1}x}$关于 x。

解决方案：

We have, 

=> $y=x^{cos^{-1}x}$

On taking log of both the sides, we get,

=> log y = log $x^{cos^{-1}x}$

=> log y = cos⁻¹ x log x

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(cos^{−1} x log x)$

=> $\frac{1}{y}\frac{dy}{dx}=cos^{-1}x(\frac{1}{x})+logx(\frac{-1}{\sqrt{1-x^2}})$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}$

=> $\frac{dy}{dx}=y\left[\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}\right]$

=> $\frac{dy}{dx}=x^{cos^{-1}x}\left[\frac{cos^{-1}x}{x}-\frac{logx}{\sqrt{1-x^2}}\right]$

问题 5. 对 x 求微分 y = (log x) ^x 。

解决方案：

We have, 

=> y = (log x)^x

On taking log of both the sides, we get,

=> log y = log (log x)^x

=> log y = x log (log x)

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (log x)]$

=> $\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{logx})(\frac{1}{x})+log(logx)$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{logx}+log(logx)$

=> $\frac{dy}{dx}=y\left[\frac{1}{logx}+log(logx)\right]$

=> $\frac{dy}{dx}=(logx)^x\left[\frac{1}{logx}+log(logx)\right]$

问题 6. 对 x 求微分 y = (log x) ^cos x。

解决方案：

We have, 

=> y = (log x)^{cos x}

On taking log of both the sides, we get,

=> log y = log (log x)^{cos x}

=> log y = cos x log (log x)

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[cos x log (log x)]$

=> $\frac{1}{y}\frac{dy}{dx}=cosx(\frac{1}{logx})(\frac{1}{x})+log(logx)(-sinx)$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{cosx}{xlogx}-sinxlog(logx)$

=> $\frac{dy}{dx}=y\left[\frac{cosx}{xlogx}-sinxlog(logx)\right]$

=> $\frac{dy}{dx}=(logx)^{cosx}\left[\frac{cosx}{xlogx}-sinxlog(logx)\right]$

问题 7. 对 x 求微分 y = (sin x) ^cos x。

解决方案：

We have, 

=> y = (sin x)^{cos x}

On taking log of both the sides, we get,

=> log y = log (sin x)^{cos x}

=> log y = cos x log (sin x)

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[cos x log (sin x)]$

=> $\frac{1}{y}\frac{dy}{dx}=cosx(\frac{1}{sinx})(cosx)+log(sinx)(-sinx)$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{cos^2x}{sinx}-sinxlog(sinx)$

=> $\frac{1}{y}\frac{dy}{dx}=cotxcosx-sinxlog(sinx)$

=> $\frac{dy}{dx}=y\left[cotxcosx-sinxlog(sinx)\right]$

=> $\frac{dy}{dx}=(sinx)^{cosx}\left[cotxcosx-sinxlog(sinx)\right]$

问题 8. 对 x 求微分 y = e ^{x log} x。

解决方案：

We have, 

=> y=e^{x log x}

=> y = $e^{logx^x}$

=> y = x^x

On taking log of both the sides, we get,

=> log y = log x^x

=> log y = x log x

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x log x)$

=> $\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{x})+logx$

=> $\frac{1}{y}\frac{dy}{dx}=1+logx$

=> $\frac{dy}{dx}=y(1+logx)$

=> $\frac{dy}{dx}=x^{x}(1+logx)$

问题 9. 对 x 微分 y = (sin x) ^log x。

解决方案：

We have, 

=> y = (sin x)^{log x}

On taking log of both the sides, we get,

=> log y = log (sin x)^{log x}

=> log y = log x log (sin x)

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log x log (sin x)]$

=> $\frac{1}{y}\frac{dy}{dx}=logx(\frac{1}{sinx})(cosx)+log(sinx)(\frac{1}{x})$

=> $\frac{1}{y}\frac{dy}{dx}=logxcotx+\frac{log(sinx)}{x}$

=> $\frac{dy}{dx}=y\left[logxcotx+\frac{log(sinx)}{x}\right]$

=> $\frac{dy}{dx}=(sinx)^{logx}\left[logxcotx+\frac{log(sinx)}{x}\right]$

问题 10. 对 x 微分 y = 10 ^{log sin} x。

解决方案：

We have, 

=> y = 10^{log sin x}

On taking log of both the sides, we get,

=> log y = log 10^{log sin x}

=> log y = log (sin x) log 10

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log (sin x) log 10]$

=> $\frac{1}{y}\frac{dy}{dx}=log10\frac{d}{dx}[log(sinx)]$

=> $\frac{1}{y}\frac{dy}{dx}=log10(\frac{1}{sinx})(cosx)$

=> $\frac{1}{y}\frac{dy}{dx}=log10cotx$

=> $\frac{dy}{dx}=ylog10cotx$

=> $\frac{dy}{dx}=10^{logsinx}[log10cotx]$

问题 11. 对 x 微分 y = (log x) ^log x。

解决方案：

We have, 

=> y = (log x)^{log x}

On taking log of both the sides, we get,

=> log y = log (log x)^{log x}

=> log y = log x log (log x)

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[log x log (log x)]$

=> $\frac{1}{y}\frac{dy}{dx}=logx(\frac{1}{logx})(\frac{1}{x})+log(logx)(\frac{1}{x})$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}+\frac{log(logx)}{x}$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{1+log(logx)}{x}$

=> $\frac{dy}{dx}=\frac{y[1+log(logx)]}{x}$

=> $\frac{dy}{dx}=\frac{(logx)^{logx}[1+log(logx)]}{x}$

问题 12. 区分$y = 10^{10^x}$关于 x。

解决方案：

We have,

=> $y = 10^{10^x}$

On taking log of both the sides, we get,

=> log y = log $10^{10^x}$

=> log y = 10^x log 10

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(10^x log 10)$

=> $\frac{1}{y}\frac{dy}{dx}=log10(10^xlog10)$

=> $\frac{1}{y}\frac{dy}{dx}=10^x(log10)^2$

=> $\frac{dy}{dx}=y10^x(log10)^2$

=> $\frac{dy}{dx}=10^{10^x}\left[10^x(log10)^2\right]$

=> $\frac{dy}{dx}=(10^{10^x+x})(log10)^2$

问题 13. 对 x 微分 y = sin x ^x 。

解决方案：

We have, 

=> y = sin x^x 

=> sin⁻¹ y = x^x

On taking log of both the sides, we get,

=> log (sin⁻¹ y) = log x^x

=> log (sin⁻¹ y) = x log x

On differentiating both sides with respect to x, we get,

=> $(\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=\frac{d}{dx}(xlogx)$

=> $(\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=x(\frac{1}{x})+logx$

=> $(\frac{1}{sin^{−1}y})(\frac{1}{\sqrt{1-y^2}})\frac{dy}{dx}=1+logx$

=> $\frac{dy}{dx}=(1+logx)(sin^{-1}y)(\sqrt{1-y^2})$

=> $\frac{dy}{dx}=(1+logx)(sin^{-1}(sinx^x))(\sqrt{1-(sinx^x)^2})$

=> $\frac{dy}{dx}=(1+logx)(x^x)(\sqrt{1-sin^2x^x})$

=> $\frac{dy}{dx}=(1+logx)(x^x)(\sqrt{cos^2x^x})$

=> $\frac{dy}{dx}=x^xcosx^x(1+logx)$

问题 14. 对 x 微分 y = (sin ^-1 x) ^x 。

解决方案：

We have, 

=> y = (sin⁻¹x)^x

On taking log of both the sides, we get,

=> log y = (sin⁻¹x)^x

=> log y = x log (sin⁻¹x)

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[x log (sin^{−1}x)]$

=> $\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{sin^{-1}x})(\frac{1}{\sqrt{1-x^2}})+log (sin^{−1}x)$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)$

=> $\frac{dy}{dx}=y\left[\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)\right]$

=> $\frac{dy}{dx}=(sin^{-1}x)^x\left[\frac{x}{sin^{-1}x(\sqrt{1-x^2})}+log (sin^{−1}x)\right]$

问题 15. 区分$y=x^{sin^{-1}x}$关于 x。

解决方案：

We have,

=> $y=x^{sin^{-1}x}$

On taking log of both the sides, we get,

=> log y = log $x^{sin^{-1}x}$

=> log y = sin⁻¹x log x

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(sin^{−1}x log x)$

=> $\frac{1}{y}\frac{dy}{dx}=sin^{−1}x(\frac{1}{x})+logx(\frac{1}{\sqrt{1-x^2}})$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}$

=> $\frac{dy}{dx}=y\left[\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}\right]$

=> $\frac{dy}{dx}=x^{sin^{-1}x}\left[\frac{sin^{-1}x}{x}+\frac{logx}{\sqrt{1-x^2}}\right]$

问题 16. 区分$y=(tanx)^{\frac{1}{x}}$关于 x。

解决方案：

We have,

=> $y=(tanx)^{\frac{1}{x}}$

On taking log of both the sides, we get,

=> log y = log $(tanx)^{\frac{1}{x}}$

=> log y = $\frac{1}{x}log(tanx)$

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[\frac{1}{x}log(tanx)]$

=> $\frac{1}{y}\frac{dy}{dx}=(\frac{1}{x})(\frac{1}{tanx})(sec^2x)+log(tanx)(\frac{-1}{x^2})$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}$

=> $\frac{dy}{dx}=y\left[\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}\right]$

=> $\frac{dy}{dx}=(tanx)^{\frac{1}{x}}\left[\frac{sec^2x}{xtanx}-\frac{log(tanx)}{x^2}\right]$

问题 17. 区分$y=x^{tan^{-1}x}$关于 x。

解决方案：

We have,

=> $y=x^{tan^{-1}x}$

On taking log of both the sides, we get,

=> log y = log $y=x^{tan^{-1}x}$

=> log y = tan⁻¹ x log x

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(tan^{−1} x log x)$

=> $\frac{1}{y}\frac{dy}{dx}=tan^{−1}x(\frac{1}{x})+logx(\frac{1}{1+x^2})$

=> $\frac{1}{y}\frac{dy}{dx}=\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}$

=> $\frac{dy}{dx}=y\left[\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}\right]$

=> $\frac{dy}{dx}=x^{tan^{-1}x}\left[\frac{tan^{-1}x}{x}+\frac{logx}{1+x^2}\right]$

问题 18. 对 x 进行微分。

(i) y = x ^x √x

解决方案：

We have,

=> y = x^x √x

On taking log of both the sides, we get,

=> log y = log (x^x √x)

=> log y = log x^x + log √x

=> log y = x log x + $\frac{1}{2}logx$

On differentiating both sides with respect to x, we get,

=> $\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}(x log x +\frac{1}{2}logx)$

=> $\frac{1}{y}\frac{dy}{dx}=x(\frac{1}{x})+logx+(\frac{1}{2})(\frac{1}{x})$

=> $\frac{1}{y}\frac{dy}{dx}=1+logx+\frac{1}{2x}$

=> $\frac{dy}{dx}=y\left(1+logx+\frac{1}{2x}\right)$

=> $\frac{dy}{dx}=x^x\sqrt{x}\left(1+logx+\frac{1}{2x}\right)$

(二) $y=x^{sinx-cosx}+\frac{x^2-1}{x^2+1}$

解决方案：

We have,

=> $y=x^{sinx-cosx}+\frac{x^2-1}{x^2+1}$

=> $y=e^{logx^{sinx-cosx}}+\frac{x^2-1}{x^2+1}$

=> $y=e^{(sinx-cosx)logx}+\frac{x^2-1}{x^2+1}$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=\frac{d}{dx}(e^{(sinx-cosx)logx}+\frac{x^2-1}{x^2+1})$

=> $\frac{dy}{dx}=(e^{(sinx-cosx)logx})[(sinx-cosx)\frac{1}{x}+logx(cosx+sinx)]+\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}$

=> $\frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{2x(x^2+1-x^2+1)}{(x^2+1)^2}$

=> $\frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{2x(2)}{(x^2+1)^2}$

=> $\frac{dy}{dx}=(x^{sinx-cosx})[\frac{sinx-cosx}{x}+logx(cosx+sinx)]+\frac{4x}{(x^2+1)^2}$

㈢ $y=x^{xcosx}+\frac{x^2+1}{x^2-1}$

解决方案：

We have,

=> $y=x^{xcosx}+\frac{x^2+1}{x^2-1}$

=> $y=e^{logx^{xcosx}}+\frac{x^2+1}{x^2-1}$

=> $y=e^{xcosxlogx}+\frac{x^2+1}{x^2-1}$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=(e^{xcosxlogx})[x((-sinx)logx+cosx(\frac{1}{x}))+cosxlogx]+\frac{(x^2-1)(2x)-(x^2+1)(2x)}{(x^2-1)^2}$

=> $\frac{dy}{dx}=(e^{xcosxlogx})[-xsinxlogx+cosx+cosxlogx]+\frac{2x(x^2-1-x^2-1)}{(x^2-1)^2}$

=> $\frac{dy}{dx}=(e^{xcosxlogx})[-xsinxlogx+cosx(1+logx)]+\frac{2x(-2)}{(x^2-1)^2}$

=> $\frac{dy}{dx}=(e^{xcosxlogx})[cosx(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1)^2}$

=> $\frac{dy}{dx}=x^{xcosx}[cosx(1+logx)-xsinxlogx]-\frac{4x}{(x^2-1)^2}$

(iv) y = (x cos x) ^x + (x sin x) ^1/x

解决方案：

We have,

=> y=(x cos x)^x + (x sin x)^1/x

=> $y=e^{log(x cos x)^x}+ e^{log(x sin x)^{\frac{1}{x}}}$

=> $y=e^{xlog(x cos x)}+ e^{\frac{1}{x}log(x sin x)}$

=> $y=e^{x(logx+logcosx)}+ e^{\frac{1}{x}(logx+logsin x)}$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=(e^{x(logx+logcosx)})[x(\frac{1}{x}+(\frac{1}{cosx})(-sinx))+log(xcosx)(1)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1}{x}(\frac{1}{x}+(\frac{1}{sinx})(cosx))+log(xsinx)(\frac{-1}{x^2})]$

=> $\frac{dy}{dx}=(e^{x(logx+logcosx)})[1-xtanx+log(xcosx)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1}{x^2}+\frac{1}{xcotx}-\frac{log(xsinx)}{x^2}]$

=> $\frac{dy}{dx}=(e^{x(logx+logcosx)})[1-xtanx+log(xcosx)]+ (e^{\frac{1}{x}(logx+logsin x)})[\frac{1-log(xsinx)+xcotx}{x^2}]$

=> $\frac{dy}{dx}=(xcosx)^x[1-xtanx+log(xcosx)]+ (xsinx)^{\frac{1}{x}}[\frac{1-log(xsinx)+xcotx}{x^2}]$

(五) $y=(x+\frac{1}{x})^x+x^{(1+\frac{1}{x})}$

解决方案：

We have,

=> $y=(x+\frac{1}{x})^x+x^{(1+\frac{1}{x})}$

=> $y=e^{log(x+\frac{1}{x})^x}+e^{logx^{(1+\frac{1}{x})}}$

=> $y=e^{xlog(x+\frac{1}{x})}+e^{(1+\frac{1}{x})logx}$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=(e^{xlog(x+\frac{1}{x})})[x(\frac{1}{x+\frac{1}{x}})(1-\frac{1}{x^2})+log(x+\frac{1}{x})]+(e^{(1+\frac{1}{x})logx})[(1+\frac{1}{x})(\frac{1}{x})+logx(\frac{-1}{x^2})]$

=> $\frac{dy}{dx}=(e^{xlog(x+\frac{1}{x})})[(\frac{x-\frac{1}{x}}{x+\frac{1}{x}})+log(x+\frac{1}{x})]+(e^{(1+\frac{1}{x})logx})[\frac{1}{x}+\frac{1}{x^2}-logx(\frac{1}{x^2})]$

=> $\frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}[\frac{1}{x}+\frac{1}{x^2}-logx(\frac{1}{x^2})]$

=> $\frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}[\frac{1}{x}+\frac{1}{x^2}-\frac{logx}{x^2}]$

=> $\frac{dy}{dx}=(x+\frac{1}{x})^x[\frac{x^2-1}{x^2+1}+log(x+\frac{1}{x})]+x^{1+\frac{1}{x}}(\frac{x+1-logx}{x^2})$

(vi) y = e ^{sin x} + (tan x) ^x

解决方案：

We have, 

=> y = e^{sin x} + (tan x)^x 

=> $y = e^{sin x} + e^{log(tanx)^x}$

=> $y = e^{sin x} + e^{xlog(tanx)}$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=\frac{d}{dx}(e^{sin x} + e^{xlog(tanx)})$

=> $\frac{dy}{dx}=(e^{sin x})(cosx) + (e^{xlogtanx})\left[x(\frac{1}{tanx})(sec^2x)+log(tanx)\right]$

=> $\frac{dy}{dx}=e^{sin x}cosx+(tanx)^x\left[\frac{xsec^2x}{tanx}+log(tanx)\right]$

(vii) y = (cos x) ^x + (sin x) ^1/x

解决方案：

We have,

=> y = (cos x)^x + (sin x)^1/x

=> $y = e^{log(cos x)^x} + e^{log(sin x)^{1/x}}$

=> $y = e^{xlog(cos x)} + e^{\frac{1}{x}log(sin x)}$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=(e^{xlog(cos x)})[x(\frac{1}{cosx})(-sinx)+log(cosx)] + (e^{\frac{1}{x}log(sin x)})[\frac{1}{x}(\frac{1}{sinx}(cosx))+log(sinx)(-\frac{1}{x^2})]$

=> $\frac{dy}{dx}=(e^{xlog(cos x)})[-xtanx+log(cosx)] + (e^{\frac{1}{x}log(sin x)})[\frac{cotx}{x}-\frac{log(sinx)}{x^2}]$

=> $\frac{dy}{dx}=(cosx)^x[-xtanx+log(cosx)] + (sinx)^{\frac{1}{x}}[\frac{cotx}{x}-\frac{log(sinx)}{x^2}]$

(八) $y=x^{x^2-3}+(x-3)^{x^2}$ , 对于 x > 3

解决方案：

We have,

=> $y=x^{x^2-3}+(x-3)^{x^2}$

=> $y=e^{logx^{x^2-3}}+e^{log(x-3)^{x^2}}$

=> $y=e^{(x^2-3)logx}+e^{x^2log(x-3)}$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=(e^{(x^2-3)logx})[(x^2-3)(\frac{1}{x})+logx(2x)]+(e^{x^2log(x-3)})[x^2(\frac{1}{x-3})+log(x-3)(2x)]$

=> $\frac{dy}{dx}=(e^{(x^2-3)logx})[\frac{x^2-3}{x}+2xlogx]+(e^{x^2log(x-3)})[\frac{x^2}{x-3}+2xlog(x-3)]$

=> $\frac{dy}{dx}=x^{x^2-3}[\frac{x^2-3}{x}+2xlogx]+(x-3)^{x^2}[\frac{x^2}{x-3}+2xlog(x-3)]$

问题 19. 当 y = e ^x + 10 ^x + x ^x时求 dy/dx。

解决方案：

We have, 

=> y = e^x + 10^x + x^x

=> $y = e^x + 10^x + e^{logx^x}$

=> $y = e^x + 10^x + e^{xlogx}$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=\frac{d}{dx}(e^x + 10^x + e^{xlogx})$

=> $\frac{dy}{dx}=e^x+10^xlog10+e^{xlogx}[x(\frac{1}{x})+logx]$

=> $\frac{dy}{dx}=e^x+10^xlog10+x^x(1+logx)$

=> $\frac{dy}{dx}=e^x+10^xlog10+x^x(logex)$

问题 20. 当 y = x ⁿ + n ^x + x ^x + n ⁿ时求 dy/dx。

解决方案：

We have, 

=> y = xⁿ + n^x + x^x + nⁿ

=> $y=x^n + n^x + e^{logx^x} + n^n$

=> $y=x^n + n^x + e^{xlogx} + n^n$

On differentiating both sides with respect to x, we get,

=> $\frac{dy}{dx}=\frac{d}{dx}(x^n+n^x+e^{xlogx}+n^n)$

=> $\frac{dy}{dx}=nx^{n-1}+n^xlogn+e^{xlogx}[x(\frac{1}{x})+logx]+0$

=> $\frac{dy}{dx}=nx^{n-1}+n^xlogn+x^x(1+logx)$

=> $\frac{dy}{dx}=nx^{n-1}+n^xlogn+x^x(logex)$