Differentiate i. y=(2x^2+lnsqrt (x))^6(1+2xsec2x)^3 cos(x+y)sin(x-y)=20x^2 lim (x^3-1)/(x^2) ii) lim ((vert x-3vert )/(x))

Let's go through each part of the question step by step.<br /><br />### Part i: Differentiation<br /><br />#### 1. Differentiate \( y = (2x^2 + \ln\sqrt{x})^6 (1 + 2x \sec(2x))^3 \)<br /><br />To differentiate \( y \), we will use the product rule and the chain rule.<br /><br />Let \( u = (2x^2 + \ln\sqrt{x})^6 \) and \( v = (1 + 2x \sec(2x))^3 \).<br /><br />The product rule states that:<br />\[ \frac{d}{dx}[uv] = u'v + uv' \]<br /><br />First, let's find \( u' \) and \( v' \).<br /><br />For \( u \):<br />\[ u = (2x^2 + \ln\sqrt{x})^6 \]<br />Let \( g(x) = 2x^2 + \ln\sqrt{x} \), then \( u = [g(x)]^6 \).<br /><br />Using the chain rule:<br />\[ u' = 6[g(x)]^5 \cdot g'(x) \]<br /><br />Now, find \( g'(x) \):<br />\[ g(x) = 2x^2 + \ln\sqrt{x} \]<br />\[ g'(x) = 4x + \frac{1}{2\sqrt{x}} \]<br /><br />So,<br />\[ u' = 6(2x^2 + \ln\sqrt{x})^5 \cdot \left(4x + \frac{1}{2\sqrt{x}}\right) \]<br /><br />For \( v \):<br />\[ v = (1 + 2x \sec(2x))^3 \]<br />Let \( h(x) = 1 + 2x \sec(2x) \), then \( v = [h(x)]^3 \).<br /><br />Using the chain rule:<br />\[ v' = 3[h(x)]^2 \cdot h'(x) \]<br /><br />Now, find \( h'(x) \):<br />\[ h(x) = 1 + 2x \sec(2x) \]<br />\[ h'(x) = 2 \sec(2x) + 4x \sec(2x) \tan(2x) \]<br /><br />So,<br />\[ v' = 3(1 + 2x \sec(2x))^2 \cdot (2 \sec(2x) + 4x \sec(2x) \tan(2x)) \]<br /><br />Now apply the product rule:<br />\[ \frac{dy}{dx} = u'v + uv' \]<br /><br />Substitute \( u \), \( u' \), \( v \), and \( v' \) into the product rule formula.<br /><br />### 2. Differentiate \( \cos(x+y) \sin(x-y) = 20x^2 \)<br /><br />To differentiate this equation with respect to \( x \), we will use the chain rule and the product rule.<br /><br />Let \( z = x + y \) and \( w = x - y \).<br /><br />Then,<br />\[ \cos(x+y) \sin(x-y) = 20x^2 \]<br />\[ \cos(z) \sin(w) = 20x^2 \]<br /><br />Differentiate both sides with respect to \( x \):<br />\[ \frac{d}{dx}[\cos(z) \sin(w)] = \frac{d}{dx}[20x^2] \]<br /><br />Using the product rule:<br />\[ \frac{d}{dx}[\cos(z) \sin(w)] = \cos(z) \cdot \frac{d}{dx}[\sin(w)] + \sin(w) \cdot \frac{d}{dx}[\cos(z)] \]<br /><br />We know:<br />\[ \frac{d}{dx}[\sin(w)] = \cos(w) \cdot \frac{dw}{dx} \]<br />\[ \frac{d}{dx}[\cos(z)] = -\sin(z) \cdot \frac{dz}{dx} \]<br /><br />So,<br />\[ \cos(z) \cdot \cos(w) \cdot \frac{dw}{dx} + \sin(w) \cdot (-\sin(z) \cdot \frac{dz}{dx}) = 40x \]<br /><br />Since \( z = x + y \) and \( w = x - y \):<br />\[ \frac{dw}{dx} = 1 - \frac{dy}{dx} \]<br />\[ \frac{dz}{dx} = 1 + \frac{dy}{dx} \]<br /><br />Substitute these into the differentiated equation:<br />\[ \cos(x+y) \cos(x-y) \cdot (1 - \frac{dy}{dx}) + \sin(x-y) \cdot (-\sin(x+y) \cdot (1 + \frac{dy}{dx})) = 40x \]<br /><br />Simplify to solve for \( \frac{dy}{dx} \).<br /><br />### 3