1. Find the fourth derivative of the function $f(x) = \ln \sqrt[3]{x^2} - \frac{42}{\sqrt[4]{x}} + \sinh(\sqrt[5]{x})$. Step 1: Simplify each term. We have $\ln \sqrt[3]{x^2} = \frac{1}{3} \ln(x^2) = \frac{2}{3} \ln x$. Also, $\frac{42}{\sqrt[4]{x}} = 42 x^{-1/4}$. And, $\sinh(\sqrt[5]{x}) = \sinh(x^{1/5})$. So, rewritten: $$ f(x) = \frac{2}{3} \ln x - 42 x^{-1/4} + \sinh(x^{1/5}) $$ Step 2: Compute derivatives term-by-term. Term 1: $\frac{2}{3} \ln x$. - 1st derivative: $\frac{2}{3} \cdot \frac{1}{x} = \frac{2}{3} x^{-1}$. - 2nd derivative: $\frac{d}{dx} \left( \frac{2}{3} x^{-1} \right) = \frac{2}{3} (-1) x^{-2} = -\frac{2}{3} x^{-2}$. - 3rd derivative: $\frac{d}{dx} \left( -\frac{2}{3} x^{-2} \right) = -\frac{2}{3} (-2) x^{-3} = \frac{4}{3} x^{-3}$. - 4th derivative: $\frac{d}{dx} \left( \frac{4}{3} x^{-3} \right) = \frac{4}{3} (-3) x^{-4} = -4 x^{-4}$. Term 2: $-42 x^{-1/4}$. - 1st derivative: $-42 \cdot (-\frac{1}{4}) x^{-5/4} = 10.5 x^{-5/4}$. - 2nd derivative: $10.5 \cdot (-\frac{5}{4}) x^{-9/4} = -\frac{52.5}{4} x^{-9/4} = -13.125 x^{-9/4}$. - 3rd derivative: $-13.125 \cdot (-\frac{9}{4}) x^{-13/4} = 29.53125 x^{-13/4}$. - 4th derivative: $29.53125 \cdot (-\frac{13}{4}) x^{-17/4} = -95.865234375 x^{-17/4}$. Term 3: $\sinh(x^{1/5})$. Let $g(x) = x^{1/5}$. - 1st derivative: $$ \frac{d}{dx} \sinh(g) = \cosh(g) \cdot g'(x) = \cosh(x^{1/5}) \cdot \frac{1}{5} x^{-4/5} $$ - 2nd derivative: $$ \frac{d}{dx} \left( \cosh(x^{1/5}) \frac{1}{5} x^{-4/5} \right) = \frac{1}{5} \left[ \sinh(x^{1/5}) \cdot g'(x) x^{-4/5} + \cosh(x^{1/5}) \cdot \frac{d}{dx} x^{-4/5} \right] $$ Calculate derivative inside: $g'(x) = \frac{1}{5} x^{-4/5}$ $\frac{d}{dx} x^{-4/5} = -\frac{4}{5} x^{-9/5}$ So, $$ f''_3 = \frac{1}{5} \left[ \sinh(x^{1/5}) \cdot \frac{1}{5} x^{-4/5} + \cosh(x^{1/5}) (-\frac{4}{5} x^{-9/5}) \right] = \frac{1}{25} \sinh(x^{1/5}) x^{-4/5} - \frac{4}{25} \cosh(x^{1/5}) x^{-9/5} $$ - 3rd derivative: Differentiate term by term using product and chain rules. Let $A = \frac{1}{25} \sinh(x^{1/5}) x^{-4/5}$ and $B = -\frac{4}{25} \cosh(x^{1/5}) x^{-9/5}$. For $A$: $$ A' = \frac{1}{25} \left[ \cosh(x^{1/5}) \cdot g'(x) x^{-4/5} + \sinh(x^{1/5}) \cdot \frac{d}{dx} x^{-4/5} \right] = \frac{1}{25} \left[ \cosh(x^{1/5}) \cdot \frac{1}{5} x^{-4/5} x^{-4/5} + \sinh(x^{1/5}) \cdot (-\frac{4}{5} x^{-9/5}) \right] = \frac{1}{125} \cosh(x^{1/5}) x^{-8/5} - \frac{4}{125} \sinh(x^{1/5}) x^{-9/5} $$ For $B$: $$ B' = -\frac{4}{25} \left[ \sinh(x^{1/5}) \cdot g'(x) x^{-9/5} + \cosh(x^{1/5}) \cdot \frac{d}{dx} x^{-9/5} \right] = -\frac{4}{25} \left[ \sinh(x^{1/5}) \cdot \frac{1}{5} x^{-4/5} x^{-9/5} + \cosh(x^{1/5}) (-\frac{9}{5} x^{-14/5}) \right] = -\frac{4}{125} \sinh(x^{1/5}) x^{-13/5} + \frac{36}{125} \cosh(x^{1/5}) x^{-14/5} $$ Finally, $$ f'''_3 = A' + B' = \frac{1}{125} \cosh(x^{1/5}) x^{-8/5} - \frac{4}{125} \sinh(x^{1/5}) x^{-9/5} - \frac{4}{125} \sinh(x^{1/5}) x^{-13/5} + \frac{36}{125} \cosh(x^{1/5}) x^{-14/5} $$ - 4th derivative: Repeat differentiation for each term. Term $\frac{1}{125} \cosh(x^{1/5}) x^{-8/5}$: $$ \frac{d}{dx} = \frac{1}{125} \left[ \sinh(x^{1/5}) \cdot g'(x) x^{-8/5} + \cosh(x^{1/5}) \frac{d}{dx} x^{-8/5} \right] $$ $g'(x) = \frac{1}{5} x^{-4/5}$ $\frac{d}{dx} x^{-8/5} = -\frac{8}{5} x^{-13/5}$ So derivative is: $$ \frac{1}{125} \left[ \sinh(x^{1/5}) \cdot \frac{1}{5} x^{-4/5} x^{-8/5} + \cosh(x^{1/5}) (-\frac{8}{5} x^{-13/5}) \right] = \frac{1}{625} \sinh(x^{1/5}) x^{-12/5} - \frac{8}{625} \cosh(x^{1/5}) x^{-13/5} $$ Term $- \frac{4}{125} \sinh(x^{1/5}) x^{-9/5}$: $$ \frac{d}{dx} = -\frac{4}{125} \left[ \cosh(x^{1/5}) g'(x) x^{-9/5} + \sinh(x^{1/5}) \frac{d}{dx} x^{-9/5} \right] = -\frac{4}{125} \left[ \cosh(x^{1/5}) \frac{1}{5} x^{-4/5} x^{-9/5} + \sinh(x^{1/5}) (-\frac{9}{5} x^{-14/5}) \right] $$ $$ = -\frac{4}{625} \cosh(x^{1/5}) x^{-13/5} + \frac{36}{625} \sinh(x^{1/5}) x^{-14/5} $$ Term $-\frac{4}{125} \sinh(x^{1/5}) x^{-13/5}$: $$ \frac{d}{dx} = -\frac{4}{125} \left[ \cosh(x^{1/5}) g'(x) x^{-13/5} + \sinh(x^{1/5}) \frac{d}{dx} x^{-13/5} \right] = -\frac{4}{125} \left[ \cosh(x^{1/5}) \frac{1}{5} x^{-4/5} x^{-13/5} + \sinh(x^{1/5}) (-\frac{13}{5} x^{-18/5}) \right] $$ $$ = -\frac{4}{625} \cosh(x^{1/5}) x^{-17/5} + \frac{52}{625} \sinh(x^{1/5}) x^{-18/5} $$ Term $\frac{36}{125} \cosh(x^{1/5}) x^{-14/5}$: $$ \frac{d}{dx} = \frac{36}{125} \left[ \sinh(x^{1/5}) g'(x) x^{-14/5} + \cosh(x^{1/5}) \frac{d}{dx} x^{-14/5} \right] = \frac{36}{125} \left[ \sinh(x^{1/5}) \frac{1}{5} x^{-4/5} x^{-14/5} + \cosh(x^{1/5}) (-\frac{14}{5} x^{-19/5}) \right] $$ $$ = \frac{36}{625} \sinh(x^{1/5}) x^{-18/5} - \frac{504}{625} \cosh(x^{1/5}) x^{-19/5} $$ Summing all four: $$ f''''_3 = \left( \frac{1}{625} \sinh(x^{1/5}) x^{-12/5} - \frac{8}{625} \cosh(x^{1/5}) x^{-13/5} \right) + \left(-\frac{4}{625} \cosh(x^{1/5}) x^{-13/5} + \frac{36}{625} \sinh(x^{1/5}) x^{-14/5} \right) + \left(-\frac{4}{625} \cosh(x^{1/5}) x^{-17/5} + \frac{52}{625} \sinh(x^{1/5}) x^{-18/5} \right) + \left( \frac{36}{625} \sinh(x^{1/5}) x^{-18/5} - \frac{504}{625} \cosh(x^{1/5}) x^{-19/5} \right) $$ Simplifying: $$ f''''_3 = \frac{1}{625} \sinh(x^{1/5}) x^{-12/5} + \frac{36}{625} \sinh(x^{1/5}) x^{-14/5} + \frac{104}{625} \sinh(x^{1/5}) x^{-18/5} - \left( \frac{8}{625} + \frac{4}{625} \right) \cosh(x^{1/5}) x^{-13/5} - \frac{4}{625} \cosh(x^{1/5}) x^{-17/5} - \frac{504}{625} \cosh(x^{1/5}) x^{-19/5} $$ $$ = \frac{1}{625} \sinh(x^{1/5}) x^{-12/5} + \frac{36}{625} \sinh(x^{1/5}) x^{-14/5} + \frac{104}{625} \sinh(x^{1/5}) x^{-18/5} - \frac{12}{625} \cosh(x^{1/5}) x^{-13/5} - \frac{4}{625} \cosh(x^{1/5}) x^{-17/5} - \frac{504}{625} \cosh(x^{1/5}) x^{-19/5} $$ Step 3: Combine all fourth derivatives: $$ f^{(4)}(x) = -4 x^{-4} - 95.865234375 x^{-17/4} + f''''_3 $$ --- 2. Show that for $F(x) = \frac{1}{1+\sin x}$, $$ (1 + \sin x)^3 \cdot F''(x) = \cos^2 x + \sin x + 1 $$ Step 1: Compute $F'(x)$: $$ F(x) = (1+\sin x)^{-1} $$ $$ F'(x) = -1 \cdot (1+\sin x)^{-2} \cdot \cos x = -\frac{\cos x}{(1+\sin x)^2} $$ Step 2: Compute $F''(x)$ by differentiating $F'(x)$ using quotient/product and chain rules: $$ F''(x) = - \frac{d}{dx} \left( \frac{\cos x}{(1+\sin x)^2} \right) = - \left[ \frac{-\sin x (1+\sin x)^2 - \cos x \cdot 2 (1+\sin x) \cdot \cos x}{(1+\sin x)^4} \right] $$ Simplify numerator: $$ -\sin x (1+\sin x)^2 - 2 \cos^2 x (1+\sin x) $$ So, $$ F''(x) = \frac{ \sin x (1+\sin x)^2 + 2 \cos^2 x (1+\sin x) }{ (1+\sin x)^4 } $$ Step 3: Multiply both sides by $(1+\sin x)^3$: $$ (1 + \sin x)^3 F''(x) = \frac{ \sin x (1+\sin x)^2 + 2 \cos^2 x (1+\sin x) }{ (1+\sin x)^4 } \cdot (1+\sin x)^3 = \frac{ \sin x (1+\sin x)^2 + 2 \cos^2 x (1+\sin x)}{1+\sin x} $$ Simplify denominator: $$ = \sin x (1+\sin x) + 2 \cos^2 x $$ Step 4: Expand $\sin x (1+\sin x)$: $$ \sin x + \sin^2 x + 2 \cos^2 x $$ Recall $\sin^2 x + \cos^2 x = 1$, so $$ \sin x + \sin^2 x + 2 \cos^2 x = \sin x + (1 - \cos^2 x) + 2 \cos^2 x = \sin x + 1 + \cos^2 x $$ This matches the RHS. --- 3. Given $x + y = \tan y$, prove that $$ y'' = -2[\cot^5 y + \cot^3 y] $$ Step 1: Differentiate the implicit equation twice. Given: $$ x + y = \tan y $$ Differentiate both sides w.r.t. $x$: $$ 1 + y' = \sec^2 y \cdot y' $$ Rearranged: $$ 1 = y' (\sec^2 y - 1) $$ Recall $\sec^2 y - 1 = \tan^2 y$, so $$ y' = \frac{1}{\tan^2 y} = \cot^2 y $$ Step 2: Differentiate $y'$ to find $y''$: $$ y'' = \frac{d}{dx} \cot^2 y = 2 \cot y \cdot \frac{d}{dx} \cot y $$ Use chain rule: $$ \frac{d}{dx} \cot y = -\csc^2 y \cdot y' $$ Substitute $y' = \cot^2 y$: $$ y'' = 2 \cot y (-\csc^2 y)(\cot^2 y) = -2 \cot y \csc^2 y \cot^2 y = -2 \cot^3 y \csc^2 y $$ Step 3: Express $\csc^2 y$ in terms of $\cot y$: Recall: $$ \csc^2 y = 1 + \cot^2 y $$ So $$ y'' = -2 \cot^3 y (1 + \cot^2 y) = -2 (\cot^3 y + \cot^5 y) $$ This matches the required formula. --- Final answers: 1. $$ f^{(4)}(x) = -4 x^{-4} - 95.865234375 x^{-17/4} + \frac{1}{625} \sinh(x^{1/5}) x^{-12/5} + \frac{36}{625} \sinh(x^{1/5}) x^{-14/5} + \frac{104}{625} \sinh(x^{1/5}) x^{-18/5} - \frac{12}{625} \cosh(x^{1/5}) x^{-13/5} - \frac{4}{625} \cosh(x^{1/5}) x^{-17/5} - \frac{504}{625} \cosh(x^{1/5}) x^{-19/5} $$ 2. $$ (1 + \sin x)^3 F''(x) = \cos^2 x + \sin x + 1 $$ 3. $$ y'' = -2[\cot^5 y + \cot^3 y] $$