Question

Find the gradient of the tangent to the curve y = (√x + 3)(3√x - 5) at the point where x = 1​

Answer 1

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:

1. Differentiate the function - this is to find a gradient (slope) at any points (technically function of slope)
2. Substitute x = x₁ in the derived function - you'll receive a slope at x = x₁ point.

First, derive the given function which is:

$\displaystyle{y = (\sqrt{x}+3)(3\sqrt{x}-5)$

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 $\displaystyle{y'=u'v+uv'}$

I'll be using product rule:

$\displaystyle{y' = (\sqrt{x}+3)'(3\sqrt{x}-5)+(\sqrt{x}+3)(3\sqrt{x}-5)'}$

Note that the following process will require you to have knowledge of Power Rules:

$\displaystyle{y = ax^n \to y' = nax^{n-1}}$

Hence:

$\displaystyle{y'=\dfrac{1}{2\sqrt{x}}(3\sqrt{x}-5) + (\sqrt{x}+3)\dfrac{3}{2\sqrt{x}}$

Now we know the derivative. Next, we find the slope at x = 1 which you substitute x = 1 in derived function:

$\displaystyle{y'(1)=\dfrac{1}{2\sqrt{1}}(3\sqrt{1}-5) + (\sqrt{1}+3)\dfrac{3}{2\sqrt{1}}}\\\\\displaystyle{y'(1)=\dfrac{1}{2}(3-5) + (1+3)\dfrac{3}{2}}\\\\\displaystyle{y'(1)=\dfrac{1}{2}(-2) + (4)\dfrac{3}{2}}\\\\\displaystyle{y'(1)=-1 + 2(3)}\\\\\displaystyle{y'(1)=-1 + 6}\\\\\displaystyle{y'(1)=5}$

Finally, we have found the slope or gradient at x = 1 which is 5.

Please let me know if you have any questions!