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))

Question

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))

Answer 1

Let's go through each part of the question step by step.### Part i: Differentiation#### 1. Differentiate \( y = (2x^2 + \ln\sqrt{x})^6 (1 + 2x \sec(2x))^3 \)To differentiate

, we will use the product rule and the chain rule.Let \( u = (2x^2 + \ln\sqrt{x})^6 \) and \( v = (1 + 2x \sec(2x))^3 \).The product rule states that:

First, let's find

and

.For

:

Let \( g(x) = 2x^2 + \ln\sqrt{x} \), then \( u = [g(x)]^6 \).Using the chain rule:

Now, find \( g'(x) \):

So,

For

:

Let \( h(x) = 1 + 2x \sec(2x) \), then \( v = [h(x)]^3 \).Using the chain rule:

Now, find \( h'(x) \):

So,

Now apply the product rule:

Substitute

,

, and

into the product rule formula.### 2. Differentiate \( \cos(x+y) \sin(x-y) = 20x^2 \)To differentiate this equation with respect to

, we will use the chain rule and the product rule.Let

and

.Then,

Differentiate both sides with respect to

:

Using the product rule:

We know:

So,

Since

and

:

Substitute these into the differentiated equation:

Simplify to solve for

.### 3

Question

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))

Solution

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