5 - 8. use differentiation to verify each equation.\n5. $\frac{d}{dx}(\tan^{3}x - 3\tan x +…

Explanation:

Step1: Recall the sum - difference rule of differentiation

The derivative of a sum/difference of functions (y = u\pm v\pm w) is (y'=u'\pm v'\pm w'). For (y=\tan^{3}x - 3\tan x+3x), we have (u = \tan^{3}x), (v = 3\tan x), (w = 3x).

Step2: Differentiate (u=\tan^{3}x) using the chain - rule

Let (t=\tan x), then (u = t^{3}). By the chain - rule (\frac{du}{dx}=\frac{du}{dt}\cdot\frac{dt}{dx}). (\frac{du}{dt}=3t^{2}) and (\frac{dt}{dx}=\sec^{2}x), so (\frac{d}{dx}(\tan^{3}x)=3\tan^{2}x\sec^{2}x).

Step3: Differentiate (v = 3\tan x)

(\frac{d}{dx}(3\tan x)=3\sec^{2}x).

Step4: Differentiate (w = 3x)

(\frac{d}{dx}(3x)=3).

Step5: Combine the derivatives

(\frac{d}{dx}(\tan^{3}x - 3\tan x+3x)=3\tan^{2}x\sec^{2}x-3\sec^{2}x + 3). Since (\sec^{2}x=1 + \tan^{2}x), we have: [ \begin{align*} &3\tan^{2}x(1 + \tan^{2}x)-3(1 + \tan^{2}x)+3\ =&3\tan^{2}x+3\tan^{4}x-3 - 3\tan^{2}x+3\ =&3\tan^{4}x \end{align*} ]

Explanation for problem 6:

Step1: Use the quotient - rule

The quotient - rule states that if (y=\frac{u}{v}), then (y'=\frac{u'v - uv'}{v^{2}}). For (y=\frac{x}{\sqrt{1 - x^{2}}}), (u = x), (v=(1 - x^{2})^{\frac{1}{2}}).

Step2: Find (u') and (v')

(u' = 1), and for (v=(1 - x^{2})^{\frac{1}{2}}), using the chain - rule, let (t = 1 - x^{2}), then (v=t^{\frac{1}{2}}), (\frac{dv}{dt}=\frac{1}{2}t^{-\frac{1}{2}}) and (\frac{dt}{dx}=-2x), so (v'=\frac{-2x}{2\sqrt{1 - x^{2}}}=\frac{-x}{\sqrt{1 - x^{2}}}).

Step3: Apply the quotient - rule

[ \begin{align*} \frac{d}{dx}\left(\frac{x}{\sqrt{1 - x^{2}}}\right)&=\frac{1\cdot\sqrt{1 - x^{2}}-x\cdot\frac{-x}{\sqrt{1 - x^{2}}}}{1 - x^{2}}\ &=\frac{\sqrt{1 - x^{2}}+\frac{x^{2}}{\sqrt{1 - x^{2}}}}{1 - x^{2}}\ &=\frac{\frac{1 - x^{2}+x^{2}}{\sqrt{1 - x^{2}}}}{1 - x^{2}}\ &=\frac{1}{(1 - x^{2})^{\frac{3}{2}}} \end{align*} ]

Explanation for problem 7:

Step1: Use the sum - difference rule

For (y=x^{4}-\ln(x^{4}+1)), if (u = x^{4}) and (v=\ln(x^{4}+1)), then (y'=u'-v').

Step2: Differentiate (u = x^{4})

(\frac{d}{dx}(x^{4}) = 4x^{3}).

Step3: Differentiate (v=\ln(x^{4}+1)) using the chain - rule

Let (t=x^{4}+1), then (v=\ln t), (\frac{dv}{dt}=\frac{1}{t}) and (\frac{dt}{dx}=4x^{3}), so (\frac{d}{dx}(\ln(x^{4}+1))=\frac{4x^{3}}{x^{4}+1}).

Step4: Combine the derivatives

[ \begin{align*} \frac{d}{dx}(x^{4}-\ln(x^{4}+1))&=4x^{3}-\frac{4x^{3}}{x^{4}+1}\ &=\frac{4x^{3}(x^{4}+1)-4x^{3}}{x^{4}+1}\ &=\frac{4x^{7}+4x^{3}-4x^{3}}{x^{4}+1}\ &=\frac{4x^{7}}{x^{4}+1} \end{align*} ]

Answer:

The equations in problems 5, 6, and 7 are verified through the above - shown differentiation steps.