Example 1. Find the derivative of
\[ f(x) = arcsin(3x^2) \]
Solution. Apply the Chain Rule to formula of the derivative of acrsine:
Then \[ f^{'}(x) = \frac{1}{\sqrt{1 - (3x^2)^2}} \frac{d}{dx}(3x^2) \]
\[ = \frac{6x}{\sqrt{1 - 9x^4}} \]
Example 2. Find the derivative of
\[ f(x) = arctan(ax^2 + bx + c) \]
Solution. Apply the Chain Rule to formula of the derivative of acrtangent:
Then \[ f^{'}(x) = \frac{1}{ax^2 + bx + c)^2} \frac{d}{dx}(ax^2 + bx + c) \]
\[ = \frac{2ax + b}{(1 + (ax^2 + bx + c)^2} \]
Example 3. Prove that:
\[ \int \frac{1}{\sqrt{a^2 - x^2}}~dx = arcsin\frac{x}{a} + C, a>0 \]
Proof. Apply Substitution Rule to arcsin:
Let \[ x = au, u = \frac{x}{a} \]
Then \[ dx = adu \]
It implies that \[ \int \frac{1}{\sqrt{a^2 - x^2}}~dx = \int \frac{1}{\sqrt{a^2 - a^2u^2}}~(adx) \]
\[ = \int \frac{1}{\sqrt{a^2(1 - u^2)}}~(adx) \]
\[ = \int \frac{1}{a\sqrt{1 - u^2}}~(adx) \]
\[ = \int \frac{1}{\sqrt{1 - u^2}}~dx \]
\[ = arcsin(u) + C \]
\[ = arcsin\frac{x}{a} + C \]
Example 4. Prove that:
\[ \int \frac{1}{a^2 - x^2}~dx = \frac{1}{a} arctan\frac{x}{a} + C, a>0 \]
Proof. Apply Substitution Rule to arctan:
Let \[ x = au, u = \frac{x}{a} \]
Then \[ dx = adu \]
It implies that \[ \int \frac{1}{a^2 - x^2}~dx = \int \frac{1}{a^2 - a^2u^2}~(adx) \]
\[ = \frac{1}{a^2} \int \frac{1}{1 - u^2}~(adx) \]
\[ = \frac{1}{a} \int \frac{1}{1 - u^2}~dx \]
\[ = \frac{1}{a} arctan(u) + C \]
\[ = \frac{1}{a} arctan\frac{x}{a} + C \]
ไม่มีความคิดเห็น:
แสดงความคิดเห็น