To show that [tex]\(\frac{dy}{dx} = 2(1 + y^2)\)[/tex] for [tex]\(y = \tan(2x)\)[/tex], let's go through the solution step by step:
1. Define the function:
Start with [tex]\(y = \tan(2x)\)[/tex].
2. Differentiate the function:
To find the derivative of [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex], use the chain rule. The chain rule states that if you have a composite function, [tex]\(f(g(x))\)[/tex], then its derivative is [tex]\(f'(g(x)) \cdot g'(x)\)[/tex].
3. Apply the chain rule:
Here, [tex]\(y = \tan(2x)\)[/tex] can be seen as [tex]\(f(g(x))\)[/tex], where [tex]\(f(u) = \tan(u)\)[/tex] and [tex]\(g(x) = 2x\)[/tex].
- First, find the derivative of the outer function [tex]\(f(u) = \tan(u)\)[/tex]. The derivative of [tex]\(\tan(u)\)[/tex] is [tex]\(\sec^2(u)\)[/tex].
- Now, find the derivative of the inner function [tex]\(g(x) = 2x\)[/tex]. The derivative of [tex]\(2x\)[/tex] is [tex]\(2\)[/tex].
Combining these results using the chain rule gives:
[tex]\[
\frac{d}{dx} \tan(2x) = \sec^2(2x) \cdot 2
\][/tex]
4. Simplify the derivative:
[tex]\[
\frac{dy}{dx} = 2 \sec^2(2x)
\][/tex]
5. Express [tex]\(\sec^2(2x)\)[/tex] in terms of [tex]\(y\)[/tex]:
Recall the trigonometric identity [tex]\(\sec^2(u) = 1 + \tan^2(u)\)[/tex].
- Here, since [tex]\(u = 2x\)[/tex],
[tex]\[
\sec^2(2x) = 1 + \tan^2(2x)
\][/tex]
6. Substitute [tex]\(\tan(2x)\)[/tex] with [tex]\(y\)[/tex]:
Given that [tex]\(y = \tan(2x)\)[/tex],
[tex]\[
\sec^2(2x) = 1 + y^2
\][/tex]
7. Put it all together:
Substitute [tex]\(\sec^2(2x)\)[/tex] in the derivative expression we obtained earlier:
[tex]\[
\frac{dy}{dx} = 2 \sec^2(2x)
\][/tex]
becomes
[tex]\[
\frac{dy}{dx} = 2 (1 + y^2)
\][/tex]
Therefore, we have shown that:
[tex]\[
\frac{dy}{dx} = 2(1 + y^2)
\][/tex]
