Question regarding finding the tangent line to a point on a curve

So, I have been working problems finding the derivative of polynomial equations. I am having some problems, however. The problem is as follows:

Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).

This is my working of the problem...

y= 4th root of x

y' = x ^ 1/4 = 1/4 x ^ -3/4 = m

therefore, the line tangent to this should be Y - Y1 = m(x - x1) so:

y - 1 = 1/4 x ^ -3/4(x - 1)

y - 1 = 1/4 x ^ 1/4 - 1/4x ^ -3/4

y = 1/4 x ^ 1/4 - 1/4x ^ -3/4 + 1

y = 1/4x + 1 <<<<<<

So by my calculations the line equation should be y = 1/4x + 1. However the book has the answer as y = 1/4x + 3/4. Where have I gone wrong? I don't see it.

Re: Question regarding finding the tangent line to a point on a curve

Quote:

Originally Posted by

**iAmKrizzle** Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).

I cannot read what you posted.

If $\displaystyle y=x^{\frac{1}{4}}$ then $\displaystyle y'=\frac{1}{4}x^{\frac{-3}{4}}$.

So at (1,1) the slope is $\displaystyle \frac{1}{4}$.

Tangent line is $\displaystyle (y-1)=\frac{1}{4}(x-1)$

Re: Question regarding finding the tangent line to a point on a curve

Quote:

Originally Posted by

**iAmKrizzle** So, I have been working problems finding the derivative of polynomial equations. I am having some problems, however. The problem is as follows:

Find the line tangent to the curve of Y = 4th root of X, on the point (1, 1).

This is my working of the problem...

y= 4th root of x

Okay, y= x^1/4)

Oh dear, no! If y= x^(1/4) then y' is NOT the same thing!

I know that's not what you **meant** to say, but is is what you actually said. Be careful what you write. Mathematics is very precise.

What you **meant** to say was

Quote:

y'= 1/4 x ^ -3/4 = m

therefore, the line tangent to this should be Y - Y1 = m(x - x1) so:

y - 1 = 1/4 x ^ -3/4(x - 1)

No, the slope or derivative **at** x= 1 is (1/4)1^(-3/4)= 1/4. It is a constant, not a function of x.

Quote:

y - 1 = 1/4 x ^ 1/4 - 1/4x ^ -3/4

y = 1/4 x ^ 1/4 - 1/4x ^ -3/4 + 1

y = 1/4x + 1 <<<<<<

So by my calculations the line equation should be y = 1/4x + 1. However the book has the answer as y = 1/4x + 3/4. Where have I gone wrong? I don't see it.

y- 1= (1/4)(x- 1)= (1/4)- 1/4. Adding 1 to both sides, y= (1/4)x+ 1- 1/4= (1/4)x+ 3/4.

Re: Question regarding finding the tangent line to a point on a curve

HallsofIvy and Plato,

Thanks your for your responses I see now where I went astray. With your help I've come to the correct answer also. Thank you again.