1. ## questions in derivatives

Hi all

I hve two Q I want the explaine how to solve

Q1 A) Find an equation for tangent to curve y = X^3 - 4X + 1 at the point (2,1)

(b ) What is the range of values of the curve's slope

Number ( A ) I can solve it but ( B) I face problem to solve

Q 2 derivative y = x - 3root X

please I want the explaine how to solve How to solve each one .

2. Originally Posted by r-soy
Hi all

I hve two Q I want the explaine how to solve

Q1 A) Find an equation for tangent to curve y = X^3 - 4X + 1 at the point (2,1)

(b ) What is the range of values of the curve's slope

Number ( A ) I can solve it but ( B) I face problem to solve

Q 2 derivative y = x - 3root X

please I want the explaine how to solve How to solve each one .
Dear r-soy,

First derivate the function, $y = x^3 - 4x + 1$, and substitute x=2. Hence you will know the gradient of the straight line which touches the curve at (2,1)

3. Yes , this for A

but now what about ( B )

......................................

and what about queation 2 ?

i wait you ?

4. Originally Posted by r-soy
Hi all

I hve two Q I want the explaine how to solve

Q1 A) Find an equation for tangent to curve y = X^3 - 4X + 1 at the point (2,1)

(b ) What is the range of values of the curve's slope

Number ( A ) I can solve it but ( B) I face problem to solve

Q 2 derivative y = x - 3root X

please I want the explaine how to solve How to solve each one .

Q1(b) the curve's slope is $y'$ ...

$y' = 3x^2 - 4$

the range of $y'$ is $[-4,\infty)$

Q2 $y = x - 3\sqrt{x}$

rewrite ...

$y = x - 3x^{\frac{1}{2}}$

use the power rule to find $y'$

5. Thanks a lot

How i can use the power rule to find

I try to use

y = 3X^1/2
y = 1.5X^-0.5

help me

6. Originally Posted by r-soy
Thanks a lot

How i can use the power rule to find

I try to use

y = 3X^1/2
y = 1.5X^-0.5 $\textcolor{red}{= \frac{3}{2\sqrt{x}}}$

help me
...

7. Originally Posted by r-soy
Thanks a lot

How i can use the power rule to find

I try to use

y = 3X^1/2
y = 1.5X^-0.5

help me
Dear r-soy,

The expression is $y=x-3x^{\frac{1}{2}}$. You have not considered the term 'x' when you derivate.

i.e: $\frac{dy}{dx}=\frac{d}{dx}x-\frac{d}{dx}(3x^{\frac{1}{2}})$