Find the derivative of the function. 1. (x^2 - 1)(x^2 + 1) 2. (3x + 1)/(2x - 1) 3. x^2sinx - xcosx 4. Find the derivative of tanx 5. Find the equation of the tangent line of f(x) a... Find the derivative of the function. 1. (x^2 - 1)(x^2 + 1) 2. (3x + 1)/(2x - 1) 3. x^2sinx - xcosx 4. Find the derivative of tanx 5. Find the equation of the tangent line of f(x) at the indicated point. f(x) = 2x^2 + 1/x^3 , x = 1
Understand the Problem
The question is asking for the derivatives of several functions and the equation of a tangent line at a specified point. The user is looking for step-by-step solutions for each part.
Answer
1. $$ 4x^3 $$ 2. $$ \frac{3(2x - 1) - (3x + 1)(2)}{(2x - 1)^2} $$ 3. $$ 2x \sin x + x^2 \cos x - \cos x + x \sin x $$ 4. $$ \sec^2 x $$ 5. $$ y = x + 2 $$
Answer for screen readers
- $$ \frac{d}{dx}((x^2 - 1)(x^2 + 1)) = 4x^3 $$
- $$ \frac{d}{dx}\left(\frac{3x + 1}{2x - 1}\right) = \frac{3(2x - 1) - (3x + 1)(2)}{(2x - 1)^2} $$
- $$ \frac{d}{dx}(x^2 \sin x - x \cos x) = 2x \sin x + x^2 \cos x - \cos x + x \sin x $$
- $$ \frac{d}{dx}(\tan x) = \sec^2 x $$
- The equation of the tangent line is ( y = x + 2 ).
Steps to Solve
- Finding the derivative of ( (x^2 - 1)(x^2 + 1) )
Use the product rule which states that if ( u = (x^2 - 1) ) and ( v = (x^2 + 1) ), then:
$$ \frac{d}{dx}(uv) = u'v + uv' $$
Here, ( u' = 2x ) and ( v' = 2x ). Therefore,
$$ \frac{d}{dx}((x^2 - 1)(x^2 + 1)) = (2x)(x^2 + 1) + (x^2 - 1)(2x) $$
- Finding the derivative of ( \frac{3x + 1}{2x - 1} )
Use the quotient rule which states that:
$$ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} $$
Let ( u = 3x + 1 ) and ( v = 2x - 1 ). Then ( u' = 3 ) and ( v' = 2 ):
$$ \frac{d}{dx}\left(\frac{3x + 1}{2x - 1}\right) = \frac{(3)(2x - 1) - (3x + 1)(2)}{(2x - 1)^2} $$
- Finding the derivative of ( x^2 \sin x - x \cos x )
Use the product rule on both terms. For ( u = x^2 ) and ( v = \sin x ):
$$ u' = 2x \quad \text{and} \quad v' = \cos x $$
So,
$$ \frac{d}{dx}(x^2 \sin x) = (2x)(\sin x) + (x^2)(\cos x) $$
For the term ( -x \cos x ), use product rule again:
$$ \frac{d}{dx}(-x \cos x) = -(1)(\cos x) - (x)(\sin x) $$
Combine both results to get the full derivative.
- Finding the derivative of ( \tan x )
The derivative of ( \tan x ) is known:
$$ \frac{d}{dx}(\tan x) = \sec^2 x $$
- Finding the equation of the tangent line of ( f(x) = 2x^2 + \frac{1}{x^3} ) at ( x = 1 )
First, find ( f(1) ):
$$ f(1) = 2(1)^2 + \frac{1}{(1)^3} = 2 + 1 = 3 $$
Next, find the derivative ( f'(x) ):
$$ f'(x) = \frac{d}{dx}(2x^2) + \frac{d}{dx}\left(\frac{1}{x^3}\right) = 4x - \frac{3}{x^4} $$
Evaluate at ( x = 1 ):
$$ f'(1) = 4(1) - 3 = 1 $$
The equation of the tangent line at ( (1, 3) ) using point-slope form ( y - y_1 = m(x - x_1) ):
$$ y - 3 = 1(x - 1) $$
This simplifies to:
$$ y = x + 2 $$
- $$ \frac{d}{dx}((x^2 - 1)(x^2 + 1)) = 4x^3 $$
- $$ \frac{d}{dx}\left(\frac{3x + 1}{2x - 1}\right) = \frac{3(2x - 1) - (3x + 1)(2)}{(2x - 1)^2} $$
- $$ \frac{d}{dx}(x^2 \sin x - x \cos x) = 2x \sin x + x^2 \cos x - \cos x + x \sin x $$
- $$ \frac{d}{dx}(\tan x) = \sec^2 x $$
- The equation of the tangent line is ( y = x + 2 ).
More Information
The derivatives calculated are crucial for understanding rates of change in functions, and the tangent line gives insight into the behavior of the function at a specific point. This lays the foundation for calculus applications, such as optimization and motion analysis.
Tips
- Forgetting to apply product or quotient rules properly.
- Not simplifying the derivatives after computing them.
- Confusing the order of operations when differentiating complex expressions.
AI-generated content may contain errors. Please verify critical information
