∫ calculus

1. **Problem (a):** Evaluate the integral $$\int_0^1 \frac{\ln(x)}{\sqrt{x}} \, dx$$. 2. Let us use the substitution $$x = t^2$$, then $$dx = 2t \, dt$$ and $$\sqrt{x} = t$$. Also,

1. **State the problem:** We need to find the derivative of the function $$y = \sin^{-1}\left(\cos(2x^3 - 3x - 5)\right).$$ 2. **Recall the derivative formula:** If $$y = \sin^{-1}

1. Problem statement: Find the derivative of $y = \sin^{-1}(\cos(2x^3 - 3x - 5))$. 2. Strategy: We treat this as a composition $y = f(g(h(x)))$ where $h(x)=2x^3-3x-5$, $g(t)=\cos t

1. The problem is to find the points on the curve $y = x^3 + 3x^2 - 9x + 4$ where the tangent line is horizontal. 2. A tangent line is horizontal where the derivative $y'$ is zero.

1. **State the problem:** Find the derivative of the function $$r = \sin(0^2) \cos(20)$$ with respect to the variable (assumed as $x$). 2. **Simplify the function:** Since $0^2 = 0

1. The problem asks to produce the power series for $\cos^2(2x)$ up to the term in $x^6$. 2. Recall the double-angle identity:

1. Differentiate $f(x) = x^2 + \frac{1}{x^2}$. - Rewrite $\frac{1}{x^2}$ as $x^{-2}$.

1. We are given the function $$g(t) = \frac{(1 + \sin 3t)^{-1}}{3 - 2t} = \frac{1}{(1 + \sin 3t)(3 - 2t)}$$ and asked to find its derivative with respect to $t$. 2. Rewrite functio

1. Use the definition of the derivative to find the derivative of the following functions: 1.1. Given $V(t) = \sqrt{14 + 3t}$.

1. **Problem 1:** Find the range of values for $k$ such that $f(x) = 1 - 2kx + kx^2 - x^3$ is decreasing over all real numbers. 2. **Step:** To determine when $f$ is decreasing eve

1. The problem states that the function \( f(X) = aX^3 - 2X^2 + 4 \) has an inflection point at \( x = \frac{1}{3} \). We need to find the value of \( \frac{f(1)}{f'(1)} \). 2. Sta

1. **Problem:** Use the definition of the derivative to find the derivatives of the functions: - $$V(t) = \sqrt{14 + 3t}$$

1. The problem is to find the derivative of the function $$g(t) = \frac{(1 + \sin 3t)^{-1}}{3 - 2t}$$. 2. Rewrite $g(t)$ as $$g(t) = \frac{1}{(1 + \sin 3t)(3 - 2t)}$$ for clarity.

1. المشكلة: لدينا الدالة $$f(x) = \frac{1}{\sqrt[3]{4x^2 - 5}}$$ ونريد إيجاد مشتقتها $$f'(x)$$. 2. كتابة الدالة بصيغة أسية لتسهيل الاشتقاق:

1. **Statement of the problem:** Find the derivative of the function: $$f(x) = \left(\sqrt[3]{4x^{2} - 5}\right)^{-1} = \frac{1}{(4x^{2} - 5)^{1/3}}$$

1. Let's start by stating the problem: Find the derivative of the function $$g(t) = \left(1 + \frac{\sin 3t}{3} - 2t\right)^{-1}$$ with respect to $t$.

1. Problem 1: Find the derivative of $$f(x) = (2x^3 - 8x^2 + 5)^4$$. 2. We apply the chain rule: If $$f(x) = [u(x)]^4$$, then $$f'(x) = 4[u(x)]^3 \cdot u'(x)$$.

1. **Problem 1: Find the derivative of** $y=\cos(\sin(x))$. 2. Let $f(u)=\cos(u)$ and $g(x)=\sin(x)$ so that $y=f(g(x))$.

1. We are given the function $$f(x) = \left(\frac{\sin x}{1 + \cos x}\right)^2$$ and need to find its derivative. 2. Start by using the chain rule: $$f'(x) = 2 \cdot \frac{\sin x}{

1. Stated problem: Evaluate the integral $$\int \frac{1}{\sin y} \, dy$$. 2. Rewrite the integrand using the cosecant function: $$\int \csc y \, dy$$.

1. **State the problem:** We need to evaluate the integral $$\int \frac{1}{\sin y} \, dy.$$ 2. **Rewrite the integral:** Recall that $$\frac{1}{\sin y} = \csc y,$$ so the integral