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

Explanation:

Step1: Apply product - rule for (a)

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. Let $u = 6x^{4}-3x^{2}+1$ and $v=\sqrt{x}=x^{\frac{1}{2}}$. Then $u' = 24x^{3}-6x$ and $v'=\frac{1}{2\sqrt{x}}$. So $f'(x)=(24x^{3}-6x)\sqrt{x}+(6x^{4}-3x^{2}+1)\cdot\frac{1}{2\sqrt{x}}$.

Step2: Apply quotient - rule for (b)

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = x^{4}\sin x$, $u'=4x^{3}\sin x+x^{4}\cos x$, $v=x^{2}+x + 1$, and $v'=2x + 1$. So $f'(x)=\frac{(x^{2}+x + 1)(4x^{3}\sin x+x^{4}\cos x)-x^{4}\sin x(2x + 1)}{(x^{2}+x + 1)^{2}}$.

Step3: Apply sum - rule and chain - rule for (c)

The sum - rule: if $y = u + v$, then $y'=u'+v'$. For $u=\tan^{3}x$, using the chain - rule ($y = u^{3}$, $u=\tan x$), $u' = 3\tan^{2}x\sec^{2}x$. For $v=\sin(3x)\tan(x^{3})$, using the product - rule and chain - rule, $v'=3\cos(3x)\tan(x^{3})+3x^{2}\sin(3x)\sec^{2}(x^{3})$. So $f'(x)=3\tan^{2}x\sec^{2}x+3\cos(3x)\tan(x^{3})+3x^{2}\sin(3x)\sec^{2}(x^{3})$.

Step4: Apply chain - rule for (d)

Let $u=\cos(\sin x)$, then $f(x)=\sin(u)$. The derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$, and the derivative of $u=\cos(\sin x)$ with respect to $x$ is $-\sin(\sin x)\cos x$. So $f'(x)=-\cos(\cos(\sin x))\sin(\sin x)\cos x$.

Answer:

(a) $f'(x)=(24x^{3}-6x)\sqrt{x}+(6x^{4}-3x^{2}+1)\cdot\frac{1}{2\sqrt{x}}$ (b) $f'(x)=\frac{(x^{2}+x + 1)(4x^{3}\sin x+x^{4}\cos x)-x^{4}\sin x(2x + 1)}{(x^{2}+x + 1)^{2}}$ (c) $f'(x)=3\tan^{2}x\sec^{2}x+3\cos(3x)\tan(x^{3})+3x^{2}\sin(3x)\sec^{2}(x^{3})$ (d) $f'(x)=-\cos(\cos(\sin x))\sin(\sin x)\cos x$