第 11 类 RD Sharma 解决方案 - 第 30 章衍生品 - 练习 30.4 |设置 1

问题 1. 对 x 微分 x ³ sin x。

解决方案：

We have, 

=> y = x³ sin x

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

$\frac{dy}{dx}=\frac{d}{dx}(x^3sinx)$

On using product rule, we get,

= $sinx\frac{d}{dx}(x^3)+x^3\frac{d}{dx}(sinx)$

= sinx (3x²) + x³ (cosx)

= 3x² sinx + x³ cosx

= x² (3 sinx + x cos x)

问题 2. 对 x 微分 x ³ e ^x 。

解决方案：

We have,

=> y = x³ e^x

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

$\frac{dy}{dx}=\frac{d}{dx}(x^3e^x)$

On using product rule, we get,

= $e^x\frac{d}{dx}(x^3)+x^3\frac{d}{dx}(e^x)$

= e^x (3x²) + x³ (e^x)

= 3x² e^x + x³ e^x

= x² e^x (3 + x)

问题 3. 对 x 微分 x ² e ^x log x。

解决方案：

We have,

=> y = x² e^x log x

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

$\frac{dy}{dx}=\frac{d}{dx}(x^2e^xlog x)$

On using product rule, we get,

= $e^xlogx\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(e^xlogx)$

On using product rule again in the second part of the expression, we get,

= $e^xlogx(2x)+x^2[logx\frac{d}{dx}(e^x)+e^x\frac{d}{dx}(logx)]$

= $e^xlogx(2x)+x^2(e^xlogx+\frac{e^x}{x})$

= $2xe^xlogx+x^2e^xlogx+xe^x$

= $xe^x(2logx+xlogx+1)$

问题 4. 区分 x ⁿ tan x 关于 x。

解决方案：

We have,

=> y = xⁿ tan x 

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

$\frac{dy}{dx}=\frac{d}{dx}(x^ntanx)$

On using product rule, we get,

= $tanx\frac{d}{dx}(x^n)+x^n\frac{d}{dx}(tanx)$

= $tanx(nx^{n-1})+x^n(sec^2x)$

= $nx^{n-1}tanx+x^nsec^2x$

= $x^{n-1}(ntanx+xsec^2x)$

问题 5. 将 x ⁿ log _a x 与 x 微分。

解决方案：

We have,

=> y = xⁿ log_a x

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

$\frac{dy}{dx}=\frac{d}{dx}(x^nlog_ax)$

On using product rule, we get,

= $log_ax\frac{d}{dx}(x^n)+x^n\frac{d}{dx}(log_ax)$

= $log_ax(nx^{n-1})+x^n\frac{d}{dx}(\frac{logx}{loga})$

= $log_ax(nx^{n-1})+\frac{x^n}{loga}(\frac{1}{x})$

= $nx^{n-1}log_ax+\frac{x^{n-1}}{loga}$

= $x^{n-1}\left(nlog_ax+\frac{1}{loga}\right)$

问题 6. 对 x 求微分e (x ³ +x ² +1)sinx 。

解决方案：

We have,

=> y = $(x^3+x^2+1)sinx$

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

$\frac{dy}{dx}=\frac{d}{dx}[(x^3+x^2+1)sinx]$

On using product rule, we get,

= $sinx\frac{d}{dx}(x^3+x^2+1)+(x^3+x^2+1)\frac{d}{dx}(sinx)$

= $sinx(3x^2+2x)+(x^3+x^2+1)(cosx)$

= $3x^2sinx+2xsinx+x^3cosx+x^2cosx+cosx$

问题 7. 区分 sin x cos x 关于 x。

解决方案：

We have,

=> y = sin x cos x

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

$\frac{dy}{dx}=\frac{d}{dx}(sinxcosx)$

On using product rule, we get,

= $cosx\frac{d}{dx}(sinx)+(sinx)\frac{d}{dx}(cosx)$

= cos x (cos x) − sin x (−sin x)

= cos² x − sin² x

= cos² x − (1 − cos² x)

= cos² x − 1 + cos² x

= 2 cos² x − 1

= cos 2x

问题 8. 区分$\frac{2^xcotx}{\sqrt{x}}$关于 x。

解决方案：

We have,

=> y = $\frac{2^xcotx}{\sqrt{x}}$

=> y = $2^x×cotx×x^{\frac{-1}{2}}$

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

$\frac{d}{dx}(\frac{2^xcotx}{\sqrt{x}})=\frac{d}{dx}(2^x×cotx×x^{\frac{-1}{2}})$

On using product rule, we get,

= $\frac{cotx}{\sqrt{x}}\frac{d}{dx}(2^x)+2^x\frac{d}{dx}(cotx×x^{\frac{-1}{2}})$

On using product rule again in the second part of the expression, we get,

= $\frac{2^xlog2cotx}{\sqrt{x}}+2^x[x^{\frac{-1}{2}}\frac{d}{dx}(cotx)+cotx\frac{d}{dx}(x^{\frac{-1}{2}})]$

= $\frac{2^xlog2cotx}{\sqrt{x}}+2^x\left[\frac{-cosec^2x}{\sqrt{x}}+\frac{-cotx}{2}\frac{1}{x^{\frac{3}{2}}}\right]$

= $\frac{2^xlog2cotx}{\sqrt{x}}-\frac{2^xcosec^2x}{\sqrt{x}}-\frac{2^xcotx}{2x^{\frac{3}{2}}}$

= $\frac{2^x}{\sqrt{x}}\left(log2cotx-cosec^2x-\frac{cotx}{2x}\right)$

问题 9. 对 x 微分 x ² sin x log x。

解决方案：

We have,

=> y = x² sin x log x

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

$\frac{dy}{dx}=\frac{d}{dx}(x^2sinxlogx)$

On using product rule, we get,

= $sinxlogx\frac{d}{dx}(x^2)+x^2\frac{d}{dx}(sinxlogx)$

On using product rule again in the second part of the expression, we get,

= $sinxlogx(2x)+x^2[logx\frac{d}{dx}(sinx)+sinx\frac{d}{dx}(logx)]$

= $2xsinxlogx+x^2[logxcosx+sinx(\frac{1}{x})]$

= $2xsinxlogx+x^2logxcosx+xsinx$

问题 10. 对 x 微分 x ⁵ e ^x + x ⁶ log x。

解决方案：

We have,

=> y = x⁵ e^x + x⁶ log x

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

$\frac{dy}{dx}=\frac{d}{dx}(x^5e^x+x^6log x)$

On using chain rule, we get,

= $\frac{d}{dx}(x^5e^x)+\frac{d}{dx}(x^6log x)$

On using product rule, we get,

= $e^x\frac{d}{dx}(x^5)+x^5\frac{d}{dx}(e^x)+logx\frac{d}{dx}(x^6)+x^6\frac{d}{dx}(logx)$

= $e^x(5x^4)+x^5(e^x)+logx(6x^5)+x^6(\frac{1}{x})$

= $5x^4e^x+x^5e^x+6x^5logx+x^5$

= $x^4(5e^x+xe^x+6xlogx+x)$