Question Statement

Find the derivative $\frac{dy}{dx}$ of the function:

$y=\left(x^2+\sin x\right)\sec x$

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Background and Explanation

This problem involves differentiating a product of two functions: a polynomial-trigonometric sum $(x^2+\sin x)$ and a trigonometric function $\sec x$. You can solve this using either the product rule directly, or by first rewriting $\sec x$ as $\frac{1}{\cos x}$ and applying the quotient rule.

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Solution

Method 1: Using the Product Rule

We treat $y$ as a product of two functions: $u=x^2+\sin x$ and $v=\sec x$.

Starting with: $y=\left(x^2+\sin x\right)\sec x$

Apply the product rule $\frac{d}{dx}(uv)=\frac{du}{dx}v+u\frac{dv}{dx}$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{d}{dx}\left(x^2+\sin x\right)\sec x+\left(x^2+\sin x\right)\frac{d}{dx}(\sec x)\\ & =(2x+\cos x)\sec x+\left(x^2+\sin x\right)\cdot (\sec x\tan x)\end{array}$

Now substitute $\sec x=\frac{1}{\cos x}$ and $\tan x=\frac{\sin x}{\cos x}$ to simplify:

$\begin{array}{rl}\frac{dy}{dx}& =(2x+\cos x)\frac{1}{\cos x}+\left(x^2+\sin x\right)\frac{1}{\cos x}\cdot \frac{\sin x}{\cos x}\\ & =\frac{2x+\cos x}{\cos x}+\frac{\left(x^2+\sin x\right)\sin x}{{\cos }^{2}x}\end{array}$

Combine the fractions over the common denominator ${\cos }^{2}x$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{(2x+\cos x)\cos x}{{\cos }^{2}x}+\frac{\left(x^2+\sin x\right)\sin x}{{\cos }^{2}x}\\ & =\frac{2x\cos x+{\cos }^{2}x+x^2\sin x+{\sin }^{2}x}{{\cos }^{2}x}\end{array}$

Group terms and apply the identity ${\sin }^{2}x+{\cos }^{2}x=1$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{2x\cos x+x^2\sin x+{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}\\ & =\frac{2x\cos x+x^2\sin x+1}{{\cos }^{2}x}\end{array}$

Method 2: Using the Quotient Rule (Alternative)

First, rewrite $\sec x$ as $\frac{1}{\cos x}$ to express $y$ as a quotient:

$\begin{array}{rl}y& =\left(x^2+\sin x\right)\cdot \frac{1}{\cos x}\\ & =\frac{x^2+\sin x}{\cos x}\end{array}$

Now apply the quotient rule $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$ where $u=x^2+\sin x$ and $v=\cos x$:

$\begin{array}{rl}\frac{dy}{dx}& =\frac{\cos x\cdot \frac{d}{dx}\left(x^2+\sin x\right)-\left(x^2+\sin x\right)\cdot \frac{d}{dx}(\cos x)}{{\cos }^{2}x}\\ & =\frac{\cos x(2x+\cos x)-\left(x^2+\sin x\right)(-\sin x)}{{\cos }^{2}x}\\ & =\frac{2x\cos x+{\cos }^{2}x+x^2\sin x+{\sin }^{2}x}{{\cos }^{2}x}\end{array}$

Simplify using ${\sin }^{2}x+{\cos }^{2}x=1$:

$\frac{dy}{dx}=\frac{2x\cos x+x^2\sin x+1}{{\cos }^{2}x}$

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Key Formulas or Methods Used

• Product Rule: $\frac{d}{dx}[u(x)v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x)$
• Quotient Rule: $\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right]=\frac{u^{\prime }(x)v(x)-u(x)v^{\prime }(x)}{[v(x){]}^2}$
• Derivatives of basic functions:
  • $\frac{d}{dx}(x^2)=2x$
  • $\frac{d}{dx}(\sin x)=\cos x$
  • $\frac{d}{dx}(\sec x)=\sec x\tan x$
  • $\frac{d}{dx}(\cos x)=-\sin x$
• Trigonometric identities:
  • $\sec x=\frac{1}{\cos x}$
  • $\tan x=\frac{\sin x}{\cos x}$
  • ${\sin }^{2}x+{\cos }^{2}x=1$

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Summary of Steps

1. Identify the structure: Recognize $y$ as a product $(x^2+\sin x)\cdot \sec x$
2. Choose your method:
   • Method 1: Apply product rule directly
   • Method 2: Rewrite as $\frac{x^2+\sin x}{\cos x}$ and use quotient rule
3. Differentiate components: Find derivatives of $x^2+\sin x$ (which is $2x+\cos x$) and $\sec x$ (which is $\sec x\tan x$) or $\cos x$ (which is $-\sin x$)
4. Apply the rule: Substitute into product rule or quotient rule formula
5. Simplify: Convert $\sec x$ and $\tan x$ to sine/cosine forms, find common denominators, and apply ${\sin }^{2}x+{\cos }^{2}x=1$
6. Final form: Express result as $\frac{2x\cos x+x^2\sin x+1}{{\cos }^{2}x}$ or equivalent form