Answer :
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]
