Find the best answers to your questions with the help of IDNLearn.com's expert contributors. Join our knowledgeable community and access a wealth of reliable answers to your most pressing questions.
Sagot :
Sure, let's start by differentiating each function with respect to [tex]\( x \)[/tex] step-by-step.
### Problem 6: Differentiate [tex]\(\left(4 x^2 - 7 x + 5\right) \cdot \sec x\)[/tex]
Given the function:
[tex]\[ f(x) = \left(4 x^2 - 7 x + 5\right) \cdot \sec x \][/tex]
To differentiate the product of two functions, we use the product rule:
[tex]\[ (uv)' = u'v + uv' \][/tex]
Let:
[tex]\[ u = 4 x^2 - 7 x + 5 \][/tex]
[tex]\[ v = \sec x \][/tex]
Now, find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = \frac{d}{dx}(4 x^2 - 7 x + 5) = 8 x - 7 \][/tex]
[tex]\[ v' = \frac{d}{dx}(\sec x) = \sec x \tan x \][/tex]
Applying the product rule:
[tex]\[ f'(x) = u'v + uv' \][/tex]
[tex]\[ f'(x) = (8 x - 7) \sec x + (4 x^2 - 7 x + 5) \sec x \tan x \][/tex]
So the derivative is:
[tex]\[ f'(x) = (8 x - 7) \sec x + (4 x^2 - 7 x + 5) \sec x \tan x \][/tex]
### Problem 7: Differentiate [tex]\( x^4(5 \sin x - 3 \cos x) \)[/tex]
Given the function:
[tex]\[ g(x) = x^4 (5 \sin x - 3 \cos x) \][/tex]
Again, use the product rule:
[tex]\[ (uv)' = u'v + uv' \][/tex]
Let:
[tex]\[ u = x^4 \][/tex]
[tex]\[ v = 5 \sin x - 3 \cos x \][/tex]
Find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = \frac{d}{dx}(x^4) = 4 x^3 \][/tex]
[tex]\[ v' = \frac{d}{dx}(5 \sin x - 3 \cos x) = 5 \cos x + 3 \sin x \][/tex]
Applying the product rule:
[tex]\[ g'(x) = u'v + uv' \][/tex]
[tex]\[ g'(x) = 4 x^3 (5 \sin x - 3 \cos x) + x^4 (5 \cos x + 3 \sin x) \][/tex]
Simplify:
[tex]\[ g'(x) = 4 x^3 (5 \sin x - 3 \cos x) + x^4 (5 \cos x + 3 \sin x) \][/tex]
### Problem 8: Differentiate [tex]\(\frac{\sqrt{x} + 1}{\sqrt{x} - 1}\)[/tex]
Given the function:
[tex]\[ h(x) = \frac{\sqrt{x} + 1}{\sqrt{x} - 1}\][/tex]
To differentiate the quotient of two functions, we use the quotient rule:
[tex]\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \][/tex]
Let:
[tex]\[ u = \sqrt{x} + 1 \][/tex]
[tex]\[ v = \sqrt{x} - 1 \][/tex]
Find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = \frac{d}{dx} (\sqrt{x} + 1) = \frac{1}{2\sqrt{x}} \][/tex]
[tex]\[ v' = \frac{d}{dx} (\sqrt{x} - 1) = \frac{1}{2\sqrt{x}} \][/tex]
Applying the quotient rule:
[tex]\[ h'(x) = \frac{u'v - uv'}{v^2} \][/tex]
[tex]\[ h'(x) = \frac{\left(\frac{1}{2\sqrt{x}}\right)(\sqrt{x} - 1) - (\sqrt{x} + 1)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x} - 1)^2} \][/tex]
Simplify:
[tex]\[ h'(x) = \frac{\frac{\sqrt{x} - 1 - \sqrt{x} - 1}{2\sqrt{x}}}{(\sqrt{x} - 1)^2} \][/tex]
[tex]\[ h'(x) = \frac{-2}{2\sqrt{x} (\sqrt{x} - 1)^2} \][/tex]
[tex]\[ h'(x) = \frac{-1}{\sqrt{x} (\sqrt{x} - 1)^2} \][/tex]
So the final derivative is:
[tex]\[ h'(x) = \frac{1}{2\sqrt{x}(\sqrt{x} - 1)} - \frac{\sqrt{x} + 1}{2\sqrt{x}(\sqrt{x} - 1)^2} \][/tex]
Thus, we have:
1. [tex]\((8 x - 7) \sec x + (4 x^2 - 7 x + 5) \sec x \tan x\)[/tex]
2. [tex]\(x^4 (3 \sin x + 5 \cos x) + 4 x^3 (5 \sin x - 3 \cos x)\)[/tex]
3. [tex]\(\frac{1}{2 \sqrt{x} (\sqrt{x} - 1)} - \frac{\sqrt{x} + 1}{2 \sqrt{x} (\sqrt{x} - 1)^2}\)[/tex]
### Problem 6: Differentiate [tex]\(\left(4 x^2 - 7 x + 5\right) \cdot \sec x\)[/tex]
Given the function:
[tex]\[ f(x) = \left(4 x^2 - 7 x + 5\right) \cdot \sec x \][/tex]
To differentiate the product of two functions, we use the product rule:
[tex]\[ (uv)' = u'v + uv' \][/tex]
Let:
[tex]\[ u = 4 x^2 - 7 x + 5 \][/tex]
[tex]\[ v = \sec x \][/tex]
Now, find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = \frac{d}{dx}(4 x^2 - 7 x + 5) = 8 x - 7 \][/tex]
[tex]\[ v' = \frac{d}{dx}(\sec x) = \sec x \tan x \][/tex]
Applying the product rule:
[tex]\[ f'(x) = u'v + uv' \][/tex]
[tex]\[ f'(x) = (8 x - 7) \sec x + (4 x^2 - 7 x + 5) \sec x \tan x \][/tex]
So the derivative is:
[tex]\[ f'(x) = (8 x - 7) \sec x + (4 x^2 - 7 x + 5) \sec x \tan x \][/tex]
### Problem 7: Differentiate [tex]\( x^4(5 \sin x - 3 \cos x) \)[/tex]
Given the function:
[tex]\[ g(x) = x^4 (5 \sin x - 3 \cos x) \][/tex]
Again, use the product rule:
[tex]\[ (uv)' = u'v + uv' \][/tex]
Let:
[tex]\[ u = x^4 \][/tex]
[tex]\[ v = 5 \sin x - 3 \cos x \][/tex]
Find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = \frac{d}{dx}(x^4) = 4 x^3 \][/tex]
[tex]\[ v' = \frac{d}{dx}(5 \sin x - 3 \cos x) = 5 \cos x + 3 \sin x \][/tex]
Applying the product rule:
[tex]\[ g'(x) = u'v + uv' \][/tex]
[tex]\[ g'(x) = 4 x^3 (5 \sin x - 3 \cos x) + x^4 (5 \cos x + 3 \sin x) \][/tex]
Simplify:
[tex]\[ g'(x) = 4 x^3 (5 \sin x - 3 \cos x) + x^4 (5 \cos x + 3 \sin x) \][/tex]
### Problem 8: Differentiate [tex]\(\frac{\sqrt{x} + 1}{\sqrt{x} - 1}\)[/tex]
Given the function:
[tex]\[ h(x) = \frac{\sqrt{x} + 1}{\sqrt{x} - 1}\][/tex]
To differentiate the quotient of two functions, we use the quotient rule:
[tex]\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \][/tex]
Let:
[tex]\[ u = \sqrt{x} + 1 \][/tex]
[tex]\[ v = \sqrt{x} - 1 \][/tex]
Find the derivatives of [tex]\( u \)[/tex] and [tex]\( v \)[/tex]:
[tex]\[ u' = \frac{d}{dx} (\sqrt{x} + 1) = \frac{1}{2\sqrt{x}} \][/tex]
[tex]\[ v' = \frac{d}{dx} (\sqrt{x} - 1) = \frac{1}{2\sqrt{x}} \][/tex]
Applying the quotient rule:
[tex]\[ h'(x) = \frac{u'v - uv'}{v^2} \][/tex]
[tex]\[ h'(x) = \frac{\left(\frac{1}{2\sqrt{x}}\right)(\sqrt{x} - 1) - (\sqrt{x} + 1)\left(\frac{1}{2\sqrt{x}}\right)}{(\sqrt{x} - 1)^2} \][/tex]
Simplify:
[tex]\[ h'(x) = \frac{\frac{\sqrt{x} - 1 - \sqrt{x} - 1}{2\sqrt{x}}}{(\sqrt{x} - 1)^2} \][/tex]
[tex]\[ h'(x) = \frac{-2}{2\sqrt{x} (\sqrt{x} - 1)^2} \][/tex]
[tex]\[ h'(x) = \frac{-1}{\sqrt{x} (\sqrt{x} - 1)^2} \][/tex]
So the final derivative is:
[tex]\[ h'(x) = \frac{1}{2\sqrt{x}(\sqrt{x} - 1)} - \frac{\sqrt{x} + 1}{2\sqrt{x}(\sqrt{x} - 1)^2} \][/tex]
Thus, we have:
1. [tex]\((8 x - 7) \sec x + (4 x^2 - 7 x + 5) \sec x \tan x\)[/tex]
2. [tex]\(x^4 (3 \sin x + 5 \cos x) + 4 x^3 (5 \sin x - 3 \cos x)\)[/tex]
3. [tex]\(\frac{1}{2 \sqrt{x} (\sqrt{x} - 1)} - \frac{\sqrt{x} + 1}{2 \sqrt{x} (\sqrt{x} - 1)^2}\)[/tex]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Find reliable answers at IDNLearn.com. Thanks for stopping by, and come back for more trustworthy solutions.
