differentiate the following functions.\n(a) $f(x)=(6x^{4}-3x^{2}+1)sqrt{x}$.\n(b) $f(x)=\frac{x^{4}sin…

Explanation:

Step1: Rewrite the function for (a)

First, rewrite $f(x)=(6x^{4}-3x^{2}+1)\sqrt{x}=(6x^{4}-3x^{2}+1)x^{\frac{1}{2}} = 6x^{\frac{9}{2}}-3x^{\frac{5}{2}}+x^{\frac{1}{2}}$. Then use the power - rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$. $f^\prime(x)=6\times\frac{9}{2}x^{\frac{9}{2}-1}-3\times\frac{5}{2}x^{\frac{5}{2}-1}+\frac{1}{2}x^{\frac{1}{2}-1}=27x^{\frac{7}{2}}-\frac{15}{2}x^{\frac{3}{2}}+\frac{1}{2}x^{-\frac{1}{2}}$

Step2: Use the quotient - rule for (b)

The quotient - rule is $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = x^{4}\sin x$, $u^\prime=x^{4}\cos x + 4x^{3}\sin x$ and $v=x^{2}+x + 1$, $v^\prime=2x + 1$. $f^\prime(x)=\frac{(x^{4}\cos x+4x^{3}\sin x)(x^{2}+x + 1)-(x^{4}\sin x)(2x + 1)}{(x^{2}+x + 1)^{2}}$

Step3: Use the sum - rule and chain - rule for (c)

The sum - rule is $\frac{d}{dx}(u + v)=\frac{d}{dx}u+\frac{d}{dx}v$. For $y = \tan^{3}x$, let $u=\tan x$, then $y = u^{3}$, and $\frac{dy}{dx}=3\tan^{2}x\sec^{2}x$. For $y=\sin(3x)\tan(x^{3})$, use the product - rule $\frac{d}{dx}(uv)=u^\prime v+uv^\prime$, where $u=\sin(3x)$, $u^\prime = 3\cos(3x)$ and $v=\tan(x^{3})$, $v^\prime=3x^{2}\sec^{2}(x^{3})$. So $f^\prime(x)=3\tan^{2}x\sec^{2}x+3\cos(3x)\tan(x^{3})+3x^{2}\sin(3x)\sec^{2}(x^{3})$

Step4: Use the chain - rule for (d)

Let $u=\cos(\sin x)$, then $y = \sin(u)$. $\frac{dy}{du}=\cos(u)$ and $\frac{du}{dx}=-\sin(\sin x)\cos x$. So $f^\prime(x)=\cos(\cos(\sin x))\times(-\sin(\sin x)\cos x)=-\cos x\sin(\sin x)\cos(\cos(\sin x))$

Answer:

(a) $f^\prime(x)=27x^{\frac{7}{2}}-\frac{15}{2}x^{\frac{3}{2}}+\frac{1}{2}x^{-\frac{1}{2}}$ (b) $f^\prime(x)=\frac{(x^{4}\cos x+4x^{3}\sin x)(x^{2}+x + 1)-(x^{4}\sin x)(2x + 1)}{(x^{2}+x + 1)^{2}}$ (c) $f^\prime(x)=3\tan^{2}x\sec^{2}x+3\cos(3x)\tan(x^{3})+3x^{2}\sin(3x)\sec^{2}(x^{3})$ (d) $f^\prime(x)=-\cos x\sin(\sin x)\cos(\cos(\sin x))$