Find the nth differential coefficient of sin²x.
Understand the Problem
The question is asking to find the nth differential coefficient of the function sin²x, which involves applying rules of differentiation to determine the nth derivative of this trigonometric function.
Answer
The nth differential coefficient of $\sin^2 x$ is given by: $$ \frac{d^n y}{dx^n} = 2^n \sin\left(2x + \frac{n\pi}{2}\right) $$
Answer for screen readers
The nth differential coefficient of the function $\sin^2 x$ is given by: $$ \frac{d^n y}{dx^n} = 2^n \sin\left(2x + \frac{n\pi}{2}\right) $$
Steps to Solve
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Identify the function and its form We need to find the nth derivative of the function $y = \sin^2 x$.
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Use trigonometric identity Using the identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$ simplifies the differentiation process.
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Differentiate the function Now we differentiate the function: $$ y = \frac{1 - \cos(2x)}{2} $$ The first derivative is: $$ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2} - \frac{1}{2}\cos(2x)\right) = \sin(2x) $$
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Differentiate iteratively Continue to differentiate to find the second derivative: $$ \frac{d^2y}{dx^2} = 2\cos(2x) $$ and for the third derivative: $$ \frac{d^3y}{dx^3} = -4\sin(2x) $$
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Observe the pattern Calculate more derivatives and note the pattern. The nth derivative alternates between sine and cosine functions, multiplying by powers of 2.
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Generalize the nth derivative The nth derivative can be expressed as: $$ \frac{d^n y}{dx^n} = 2^n \sin\left(2x + \frac{n\pi}{2}\right) $$
The nth differential coefficient of the function $\sin^2 x$ is given by: $$ \frac{d^n y}{dx^n} = 2^n \sin\left(2x + \frac{n\pi}{2}\right) $$
More Information
This result shows how the sine function's periodic behavior is affected by the derivatives, with each derivative introducing a factor of 2 and a phase shift. This pattern highlights the relationship between trigonometric functions and their derivatives.
Tips
- Forgetting to apply the trigonometric identity before differentiating.
- Losing track of the patterns in derivatives, especially when working with higher derivatives.
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