Answer:
Gradient (slope) is 5.
Step-by-step explanation:
To find the gradient (slope) of a curve graph at x = x₁ can be done by steps following:
- Differentiate the function - this is to find a gradient (slope) at any points (technically function of slope)
- Substitute x = x₁ in the derived function - you'll receive a slope at x = x₁ point.
First, derive the given function which is:
[tex]\displaystyle{y = (\sqrt{x}+3)(3\sqrt{x}-5)[/tex]
Differentiation can be done two ways - go ahead and expand the expression then derive it or you can use the product rule where it states that [tex]\displaystyle{y'=u'v+uv'}[/tex]
I'll be using product rule:
[tex]\displaystyle{y' = (\sqrt{x}+3)'(3\sqrt{x}-5)+(\sqrt{x}+3)(3\sqrt{x}-5)'}[/tex]
Note that the following process will require you to have knowledge of Power Rules:
[tex]\displaystyle{y = ax^n \to y' = nax^{n-1}}[/tex]
Hence:
[tex]\displaystyle{y'=\dfrac{1}{2\sqrt{x}}(3\sqrt{x}-5) + (\sqrt{x}+3)\dfrac{3}{2\sqrt{x}}[/tex]
Now we know the derivative. Next, we find the slope at x = 1 which you substitute x = 1 in derived function:
[tex]\displaystyle{y'(1)=\dfrac{1}{2\sqrt{1}}(3\sqrt{1}-5) + (\sqrt{1}+3)\dfrac{3}{2\sqrt{1}}}\\\\\displaystyle{y'(1)=\dfrac{1}{2}(3-5) + (1+3)\dfrac{3}{2}}\\\\\displaystyle{y'(1)=\dfrac{1}{2}(-2) + (4)\dfrac{3}{2}}\\\\\displaystyle{y'(1)=-1 + 2(3)}\\\\\displaystyle{y'(1)=-1 + 6}\\\\\displaystyle{y'(1)=5}[/tex]
Finally, we have found the slope or gradient at x = 1 which is 5.
Please let me know if you have any questions!
