Question

How do you find an equation of the tangent line to the graph `f(x)=2/root4(x^3)` at (1,2)?

Answer 1

`y=-3/2x+7/2`

Explanation:

Start by simplifying the expression a little.

You should know two formulae for indices (there are more, but these two are relevant):

`1/x = x^-1`

`root(a)(x^b) = x^(b/a)`

From these we can see that

`2/root(4)(x^3) = 2 * 1/root(4)(x^3)`

`= 2 * 1/x^(3/4) = 2x^(-3/4)`

Now the expression is simplified, differentiating to find the gradient is easy is much easier. (Multiply the whole thing by the power and then subtract `1` from the power to differentiate.)

`d/dx 2x^(-3/4) = -3/4*2x^(-3/4-1)`

`=-3/2x^(-7/4)`

In the question, it gives the point `(1, 2)`, so the gradient of the line (or `m` in `y=mx+b`) is found by substituting `1` (the `x` value) into the derivative.

`m=-3/2(1)^(-7/4)=-3/2`

So

`y = -3/2x + b`

To find `b`, also called `c` sometimes, we substitute in the entire point: `x=1, y=2`, so

`2 = -3/2*1+b = b-3/2`

`b=7/2`

So

`y = -3/2x+7/2`