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 $h'(x)$, t... 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 the function from first principles, we use the definition of the derivative:

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

Let (f(x) = (3x - 5)(2x + 4)), which expands to:

$f(x) = 6x^2 + 12x - 20$

Now, calculating (f(x + h)):

= 6(x^2 + 2xh + h^2) + 12x + 12h - 20\ = 6x^2 + 12xh + 6h^2 + 12x + 12h - 20$$ Now substitute into our limit: $$\lim_{h \to 0} \frac{(6x^2 + 12xh + 6h^2 + 12x + 12h - 20) - (6x^2 + 12x - 20)}{h}$$ This simplifies to: $$\lim_{h \to 0} \frac{12xh + 6h^2 + 12h}{h} = \lim_{h \to 0} (12x + 6h + 12) = 12x + 12$$ Thus, the derivative is: $$f'(x) = 12x + 12$$

Step 2

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

99%

104 rated

Answer

To find the derivative of (h(x) = \frac{1}{2}\ln(2x + 3) + C), we use the chain rule:

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

Step 3

(ii) Find the value of $A$.

96%

101 rated

Answer

To find the value of (A) from the area between the curve and the x-axis, we can set up the integral:

$\int_0^A h'(x) \, dx = \int_0^A \frac{1}{2x + 3} \, dx$

This evaluates to:

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

Setting this equal to (\ln 3):

$\frac{1}{2} \ln(2A + 3) - \frac{1}{2} \ln(3) = \ln 3$

Combine the logarithms:

$\frac{1}{2} \ln\left(\frac{2A + 3}{3}\right) = \ln 3$

Now, multiply both sides by 2:

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

This can be rewritten as:

$\frac{2A + 3}{3} = 9$

Then, multiplying both sides by 3:

$2A + 3 = 27$

Subtracting 3 gives:

$2A = 24\quad \Rightarrow \quad A = 12$

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.

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

(ii) Find the value of $A$.

Join the Leaving Cert students using SimpleStudy...