(a) Differentiate $(3x - 5)(2x + 4)$ with respect to $x$ from first principles. (b) (i) $h(x) = \frac{1}{2} \ln(2x + 3) + C$, where $C$ is a constant. Find $h'(x)$, the derivative of $h(x)$. (ii) The diagram below shows part of the graph of the function $h'(x)$. The shaded region in the diagram is between the graph and the x-axis, from $x = 0$ to $x = A$. This shaded region has an area of $ ext{ln}3$ square units. Find the value of $A$.

Photo AI

(a) Differentiate $(3x - 5)(2x + 4)$ with respect to $x$ from first principles - Leaving Cert Mathematics - Question 6 - 2020

Question 6

(a) Differentiate $(3x - 5)(2x + 4)$ with respect to $x$ from first principles. (b) (i) $h(x) = \frac{1}{2} \ln(2x + 3) + C$, where $C$ is a constant. Find $h'(x)$... show full transcript

Worked Solution & Example Answer:(a) Differentiate $(3x - 5)(2x + 4)$ with respect to $x$ from first principles - Leaving Cert Mathematics - Question 6 - 2020

Step 1

Differentiate $(3x - 5)(2x + 4)$ with respect to $x$ from first principles

96%

114 rated

Answer

To differentiate from first principles, we apply the definition of the derivative:

$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$

Here, let ( f(x) = (3x - 5)(2x + 4) ). Therefore:

Calculate ( f(x + h) ):

$f(x + h) = (3(x + h) - 5)(2(x + h) + 4)$ Expanding this gives:

$= (3x + 3h - 5)(2x + 2h + 4)$

$= (3x - 5)(2x + 4) + 6hx + 12h + 2h^2$
Then calculate ( f(x + h) - f(x) ):

$= [6x^2 + 12x - 20 + 6hx + 12h + 2h^2] - [6x^2 + 12x - 20]$

$= 6hx + 12h + 2h^2$
Now, substitute this into the limit:

$f'(x) = \lim_{h \to 0} \frac{6hx + 12h + 2h^2}{h} = \lim_{h \to 0} (6x + 12 + 2h)$

Taking the limit as ( h \to 0 ):

$f'(x) = 6x + 12$

Step 2

Find $h'(x)$, the derivative of $h(x)$

99%

104 rated

Answer

Given: $h(x) = \frac{1}{2} \ln(2x + 3) + C$

To find the derivative $h'(x)$ , we use the chain rule:

Differentiate:

$h'(x) = \frac{1}{2} \cdot \frac{1}{2x + 3} \cdot (2) = \frac{1}{2x + 3}$

Step 3

Find the value of $A$ such that the area under $h'(x)$ from $0$ to $A$ is $ ext{ln}3$

96%

101 rated

Answer

We want to find $A$ such that:

$\int_0^A h'(x) \, dx = \text{ln}3$

Substitute $h'(x)$ into the integral:

$\int_0^A \frac{1}{2x + 3} \, dx$
Solve the integral:

$= \frac{1}{2} \ln(2x + 3) \Bigg|_0^A = \frac{1}{2} \left[ \ln(2A + 3) - \ln(3) \right]$

$= \frac{1}{2} \ln \left(\frac{2A + 3}{3}\right)$
Set the equation equal to $ext{ln}3$ :

$\frac{1}{2} \ln \left(\frac{2A + 3}{3}\right) = \text{ln}3$
Multiply both sides by 2:

$\ln \left(\frac{2A + 3}{3}\right) = 2\text{ln}3$

$= \ln(9)$
Exponentiate:

$\frac{2A + 3}{3} = 9$
Solve for $A$ :

$2A + 3 = 27$

$2A = 24$

$A = 12$

(a) Differentiate $(3x - 5)(2x + 4)$ with respect to $x$ from first principles - Leaving Cert Mathematics - Question 6 - 2020

Worked Solution & Example Answer:(a) Differentiate $(3x - 5)(2x + 4)$ with respect to $x$ from first principles - Leaving Cert Mathematics - Question 6 - 2020

Differentiate $(3x - 5)(2x + 4)$ with respect to $x$ from first principles

Find $h'(x)$, the derivative of $h(x)$

Find the value of $A$ such that the area under $h'(x)$ from $0$ to $A$ is $ ext{ln}3$

Join the Leaving Cert students using SimpleStudy...