Answer :
Alright, let's break down these mathematics problems step by step.
### Problem 10
Evaluate [tex]\(\lim _{x \rightarrow \infty} \frac{2 x-1}{\sqrt{x^3-1}}\)[/tex]
Solution:
First, factor out the highest power of [tex]\(x\)[/tex] from the numerator and the denominator:
[tex]\[ \frac{2x - 1}{\sqrt{x^3 - 1}} = \frac{2x\left(1 - \frac{1}{2x}\right)}{\sqrt{x^3\left(1 - \frac{1}{x^3}\right)}} \][/tex]
Simplify inside the square root:
[tex]\[ = \frac{2x(1 - \frac{1}{2x})}{x^{3/2}\sqrt{1 - \frac{1}{x^3}}} = \frac{2(1 - \frac{1}{2x})}{x^{1/2}\sqrt{1 - \frac{1}{x^3}}} \][/tex]
As [tex]\(x \to \infty\)[/tex], [tex]\(\frac{1}{2x} \to 0\)[/tex] and [tex]\(\frac{1}{x^3} \to 0\)[/tex]:
[tex]\[ \frac{2(1 - 0)}{x^{1/2}\sqrt{1 - 0}} = \frac{2}{x^{1/2}} \][/tex]
Since [tex]\(x^{1/2} = \sqrt{x}\)[/tex] continues to infinity, the limit:
[tex]\[ \lim_{x \to \infty} \frac{2}{x^{1/2}} = 0 \][/tex]
Hence:
[tex]\[ \boxed{0} \][/tex]
### Problem 11
Evaluate [tex]\(\lim _{x \rightarrow 0} \frac{x \cos (x)-\sin (x)}{4 x^x}\)[/tex]
Solution:
Rewrite [tex]\(4x^x\)[/tex]:
[tex]\[ 4x^x = 4e^{x \ln x} \][/tex]
As [tex]\(x \to 0\)[/tex], [tex]\( x \ln x \to 0 \)[/tex]:
[tex]\[ 4 \cdot 1 = 4 \][/tex]
Use the approximation [tex]\(\cos(x) \approx 1\)[/tex] and [tex]\(\sin(x) \approx x\)[/tex] for small [tex]\(x\)[/tex]:
[tex]\[ \frac{x \cos(x) - \sin(x)}{4} = \frac{x (1) - x}{4} = \frac{0}{4} = 0 \][/tex]
Hence:
[tex]\[ \boxed{0} \][/tex]
### Problem 12
If [tex]\(y = (1-x)^{(1-x)}\)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex]
Solution:
Let [tex]\(y = (1-x)^{(1-x)}\)[/tex]. Taking natural logarithm on both sides:
[tex]\[ \ln y = (1-x)\ln(1-x) \][/tex]
Differentiate implicitly:
[tex]\[ \frac{1}{y}\frac{dy}{dx} = \ln(1-x) + (1-x)\left(-\frac{1}{1-x}\right) \][/tex]
Simplify:
[tex]\[ \frac{dy}{dx} = y \left(\ln(1-x) - 1\right) \][/tex]
Thus:
[tex]\[ \frac{dy}{dx} = (1-x)^{(1-x)}(ln(1-x) - 1) \][/tex]
### Problem 13
If [tex]\( y = (1 + x^2)^{19}\)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex]
Solution:
Use the chain rule. Let [tex]\(u = 1 + x^2\)[/tex]. Then:
[tex]\[ y = u^{19} \][/tex]
[tex]\[ \frac{dy}{dx} = 19u^{18} \cdot \frac{du}{dx} = 19(1 + x^2)^{18} \cdot 2x = 38x(1+x^2)^{18} \][/tex]
Thus:
[tex]\[ \boxed{\frac{38xy}{1+x^2}} \][/tex]
### Problem 14
If [tex]\(x = c^t + \sin (t)\)[/tex] and [tex]\(y = c^t - \cos (t)\)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex]
Solution:
Find the derivatives [tex]\(\frac{dx}{dt}\)[/tex] and [tex]\(\frac{dy}{dt}: \[ \frac{dx}{dt} = c^t \ln c + \cos(t) \] \[ \frac{dy}{dt} = c^t \ln c + \sin (t) \] Thus: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{c^t \ln c + \sin t}{c^t \ln c + \cos t} \] ### Problem 15 If \(y = e^{7x}\sin(3x)\)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex] at [tex]\(x = 0\)[/tex]
Solution:
Differentiate [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] using product rule:
[tex]\[ \frac{dy}{dx} = e^{7x}\cdot 7 \sin(3x) + e^{7x}\cdot 3\cos(3x)= e^{7x}(7 \sin(3x) + 3 \cos (3x)) \][/tex]
At [tex]\(x = 0\)[/tex]:
[tex]\[ = e^{0}(7 \sin(0) + 3 \cos(0)) = 3 \][/tex]
Thus:
[tex]\[ \boxed{3} \][/tex]
### Problem 10
Evaluate [tex]\(\lim _{x \rightarrow \infty} \frac{2 x-1}{\sqrt{x^3-1}}\)[/tex]
Solution:
First, factor out the highest power of [tex]\(x\)[/tex] from the numerator and the denominator:
[tex]\[ \frac{2x - 1}{\sqrt{x^3 - 1}} = \frac{2x\left(1 - \frac{1}{2x}\right)}{\sqrt{x^3\left(1 - \frac{1}{x^3}\right)}} \][/tex]
Simplify inside the square root:
[tex]\[ = \frac{2x(1 - \frac{1}{2x})}{x^{3/2}\sqrt{1 - \frac{1}{x^3}}} = \frac{2(1 - \frac{1}{2x})}{x^{1/2}\sqrt{1 - \frac{1}{x^3}}} \][/tex]
As [tex]\(x \to \infty\)[/tex], [tex]\(\frac{1}{2x} \to 0\)[/tex] and [tex]\(\frac{1}{x^3} \to 0\)[/tex]:
[tex]\[ \frac{2(1 - 0)}{x^{1/2}\sqrt{1 - 0}} = \frac{2}{x^{1/2}} \][/tex]
Since [tex]\(x^{1/2} = \sqrt{x}\)[/tex] continues to infinity, the limit:
[tex]\[ \lim_{x \to \infty} \frac{2}{x^{1/2}} = 0 \][/tex]
Hence:
[tex]\[ \boxed{0} \][/tex]
### Problem 11
Evaluate [tex]\(\lim _{x \rightarrow 0} \frac{x \cos (x)-\sin (x)}{4 x^x}\)[/tex]
Solution:
Rewrite [tex]\(4x^x\)[/tex]:
[tex]\[ 4x^x = 4e^{x \ln x} \][/tex]
As [tex]\(x \to 0\)[/tex], [tex]\( x \ln x \to 0 \)[/tex]:
[tex]\[ 4 \cdot 1 = 4 \][/tex]
Use the approximation [tex]\(\cos(x) \approx 1\)[/tex] and [tex]\(\sin(x) \approx x\)[/tex] for small [tex]\(x\)[/tex]:
[tex]\[ \frac{x \cos(x) - \sin(x)}{4} = \frac{x (1) - x}{4} = \frac{0}{4} = 0 \][/tex]
Hence:
[tex]\[ \boxed{0} \][/tex]
### Problem 12
If [tex]\(y = (1-x)^{(1-x)}\)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex]
Solution:
Let [tex]\(y = (1-x)^{(1-x)}\)[/tex]. Taking natural logarithm on both sides:
[tex]\[ \ln y = (1-x)\ln(1-x) \][/tex]
Differentiate implicitly:
[tex]\[ \frac{1}{y}\frac{dy}{dx} = \ln(1-x) + (1-x)\left(-\frac{1}{1-x}\right) \][/tex]
Simplify:
[tex]\[ \frac{dy}{dx} = y \left(\ln(1-x) - 1\right) \][/tex]
Thus:
[tex]\[ \frac{dy}{dx} = (1-x)^{(1-x)}(ln(1-x) - 1) \][/tex]
### Problem 13
If [tex]\( y = (1 + x^2)^{19}\)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex]
Solution:
Use the chain rule. Let [tex]\(u = 1 + x^2\)[/tex]. Then:
[tex]\[ y = u^{19} \][/tex]
[tex]\[ \frac{dy}{dx} = 19u^{18} \cdot \frac{du}{dx} = 19(1 + x^2)^{18} \cdot 2x = 38x(1+x^2)^{18} \][/tex]
Thus:
[tex]\[ \boxed{\frac{38xy}{1+x^2}} \][/tex]
### Problem 14
If [tex]\(x = c^t + \sin (t)\)[/tex] and [tex]\(y = c^t - \cos (t)\)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex]
Solution:
Find the derivatives [tex]\(\frac{dx}{dt}\)[/tex] and [tex]\(\frac{dy}{dt}: \[ \frac{dx}{dt} = c^t \ln c + \cos(t) \] \[ \frac{dy}{dt} = c^t \ln c + \sin (t) \] Thus: \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{c^t \ln c + \sin t}{c^t \ln c + \cos t} \] ### Problem 15 If \(y = e^{7x}\sin(3x)\)[/tex], find [tex]\(\frac{dy}{dx}\)[/tex] at [tex]\(x = 0\)[/tex]
Solution:
Differentiate [tex]\(y\)[/tex] with respect to [tex]\(x\)[/tex] using product rule:
[tex]\[ \frac{dy}{dx} = e^{7x}\cdot 7 \sin(3x) + e^{7x}\cdot 3\cos(3x)= e^{7x}(7 \sin(3x) + 3 \cos (3x)) \][/tex]
At [tex]\(x = 0\)[/tex]:
[tex]\[ = e^{0}(7 \sin(0) + 3 \cos(0)) = 3 \][/tex]
Thus:
[tex]\[ \boxed{3} \][/tex]
