Differentiation of Inverse Functions

Approach: Use the inverse function differentiation formula together with the chain rule.

Steps:

1. Let $y = \arcsin(x^2)$, so $x^2 = \sin y$
2. Differentiate both sides with respect to $x$: $2x = \cos y \cdot y'$
3. Solve for $y'$: $y' = \frac{2x}{\cos y} = \frac{2x}{\sqrt{1 - x^4}}$

Answer: $f'(x) = \frac{2x}{\sqrt{1 - x^4}}$

Approach: This is a composite function that involves an inverse trigonometric function and a logarithmic function.

Steps:

1. Outer layer: $f(u) = \arctan u$, $f'(u) = \frac{1}{1 + u^2}$
2. Inner layer: $g(x) = \ln x$, $g'(x) = \frac{1}{x}$
3. Derivative of the composite function: $f'(x) = \frac{1}{1 + (\ln x)^2} \cdot \frac{1}{x} = \frac{1}{x(1 + (\ln x)^2)}$

Answer: $f'(x) = \frac{1}{x(1 + (\ln x)^2)}$

Approach: This is a composite function that involves an inverse trigonometric function and a power function.

Steps:

1. Outer layer: $f(u) = \arccos u$, $f'(u) = -\frac{1}{\sqrt{1 - u^2}}$
2. Inner layer: $g(x) = \sqrt{1 - x^2}$, $g'(x) = \frac{-x}{\sqrt{1 - x^2}}$
3. Derivative of the composite function: $f'(x) = -\frac{1}{\sqrt{1 - (1 - x^2)}} \cdot \frac{-x}{\sqrt{1 - x^2}} = \frac{x}{|x|\sqrt{1 - x^2}}$

Answer: $f'(x) = \frac{x}{|x|\sqrt{1 - x^2}}$

Approach: This is a composite function that involves an inverse hyperbolic function and an exponential function.

Steps:

1. Outer layer: $f(u) = \sinh^{-1} u$, $f'(u) = \frac{1}{\sqrt{1 + u^2}}$
2. Inner layer: $g(x) = e^x$, $g'(x) = e^x$
3. Derivative of the composite function: $f'(x) = \frac{1}{\sqrt{1 + e^{2x}}} \cdot e^x = \frac{e^x}{\sqrt{1 + e^{2x}}}$

Answer: $f'(x) = \frac{e^x}{\sqrt{1 + e^{2x}}}$

Approach: This is a composite function that involves a logarithmic function and an inverse trigonometric function.

Steps:

1. Outer layer: $f(u) = \ln u$, $f'(u) = \frac{1}{u}$
2. Inner layer: $g(x) = \arccos x$, $g'(x) = -\frac{1}{\sqrt{1 - x^2}}$
3. Derivative of the composite function: $f'(x) = \frac{1}{\arccos x} \cdot (-\frac{1}{\sqrt{1 - x^2}}) = -\frac{1}{\arccos x \cdot \sqrt{1 - x^2}}$

Answer: $f'(x) = -\frac{1}{\arccos x \cdot \sqrt{1 - x^2}}$

Approach: Use the property of equivalent infinitesimals together with the idea behind differentiating inverse functions.

Steps:

1. From $\lim\limits_{x\to1} \frac{f(x)}{\ln x}=1$ we get $f(x)\sim \ln x$ as $x\to1$

2. As $x\to1$, $\ln x\to0$, so $\lim\limits_{x\to1} f(x)=0$

3. Let $g(x)=\frac{f(x)}{\ln x}$, then $f(x)=g(x)\ln x$

4. Apply the product rule: $f'(x)=g'(x)\ln x + g(x)\cdot\frac{1}{x}$

5. As $x\to1$, $\ln x\to0$, $\frac{1}{x}\to1$, $g(x)\to1$

6. Therefore $f'(1)=\lim\limits_{x\to1} f'(x)=1$

Answer: $f'(1)=1$

Approach: Use the parametric differentiation formula, which involves the idea behind differentiating inverse functions.

Steps:

1. When $t\geq0$, $x=3t, y=t\sin t$, so $y=\frac{x}{3}\sin\frac{x}{3}$

2. When $t<0$, $x=t, y=-t\sin t$, so $y=-x\sin x$

3. Hence $f(x)=\begin{cases}\frac{x}{3}\sin\frac{x}{3}, & x\geq0\\-x\sin x, & x<0\end{cases}$

4. Compute the left derivative: $f'_-(0)=\lim\limits_{x\to0^-}\frac{-x\sin x-0}{x}=-\lim\limits_{x\to0^-}\sin x=0$

5. Compute the right derivative: $f'_+(0)=\lim\limits_{x\to0^+}\frac{\frac{x}{3}\sin\frac{x}{3}-0}{x}=\lim\limits_{x\to0^+}\frac{\sin\frac{x}{3}}{3}=0$

6. Since the left and right derivatives are equal, $f'(0)=0$

Answer: $f'(0)=0$

Approach: Use the derivative formula for a function with a variable upper limit together with the differentiation of composite functions.

Steps:

1. By the derivative formula for a function with a variable upper limit: $f'(x)=e^{\cos x}$

2. Differentiate $f'(x)$ once more, using the differentiation of composite functions: let $u=\cos x$, then $f'(x)=e^u$

   $\frac{d}{dx}(e^u)=e^u\cdot\frac{du}{dx}=e^{\cos x}\cdot(-\sin x)=-e^{\cos x}\sin x$

Answer: $f'(x)=e^{\cos x}$ $f''(x)=-e^{\cos x}\sin x$

Approach: Use the derivative formula for a function with a variable upper limit together with the differentiation of composite functions.

Steps:

1. By the derivative formula for a function with a variable upper limit: $f'(x) = e^{x^2} \sin x$

2. Differentiate $f'(x)$ once more, using the product rule and the differentiation of composite functions: let $u = e^{x^2}$, $v = \sin x$

   $u' = e^{x^2} \cdot 2x = 2x e^{x^2}$ (differentiation of a composite function) $v' = \cos x$

3. Apply the product rule: $f''(x) = u'v + uv' = 2x e^{x^2} \cdot \sin x + e^{x^2} \cdot \cos x$ $= 2x e^{x^2} \sin x + e^{x^2} \cos x$

Answer: $f'(x) = e^{x^2} \sin x$ $f''(x) = 2x e^{x^2} \sin x + e^{x^2} \cos x$

Approach: Use the method for solving a first-order linear differential equation.

Steps:

1. This is a first-order linear differential equation: $y'+\frac{1}{2\sqrt{x}}y=2+\sqrt{x}$

2. Integrating factor: $\mu(x)=e^{\int\frac{1}{2\sqrt{x}}dx}=e^{\sqrt{x}}$

3. Multiply both sides of the equation by the integrating factor: $e^{\sqrt{x}}y'+e^{\sqrt{x}}\frac{1}{2\sqrt{x}}y=(2+\sqrt{x})e^{\sqrt{x}}$

4. The left-hand side can be written as: $(e^{\sqrt{x}}y)'=(2+\sqrt{x})e^{\sqrt{x}}$

5. Integrating gives: $e^{\sqrt{x}}y=\int(2+\sqrt{x})e^{\sqrt{x}}dx$

6. Evaluate the integral: $\int(2+\sqrt{x})e^{\sqrt{x}}dx=2\int e^{\sqrt{x}}dx+\int\sqrt{x}e^{\sqrt{x}}dx$

7. Let $u=\sqrt{x}$, so $dx=2udu$ and the integral becomes: $2\int e^u\cdot2udu+\int ue^u\cdot2udu=4\int ue^udu+2\int u^2e^udu$

8. Solve with integration by parts; in the end we get: $y(x)=e^{-\sqrt{x}}\left(2x e^{\sqrt{x}}+C\right)$

9. From $y(1)=3$ we get $C=e$, so: $y(x)=e^{-\sqrt{x}}\left(2x e^{\sqrt{x}}+e\right)$

Answer: $y(x)=e^{-\sqrt{x}}\left(2x e^{\sqrt{x}}+e\right)$