Need help to understand how the slope was derived in this problem

**alex95** · January 15th 2011, 03:38 PM

I am kind of struggling to understand how the slope is derived in the following problem:
"Find the equation of the tangent line to the curve y = x^2 through the point (7, 49)."

The answer is that the slope is 2x but I don't understand how they got there since there is only one set of points given.

Thank you for the help,

Alex

**dwsmith** · January 15th 2011, 03:41 PM

$y'=\mbox{exponent}*x^{\mbox{exponent -1}}$

$y'=2x^{2-1}$

**Prove It** · January 15th 2011, 03:50 PM

The slope at any point $\displaystyle x$ on the curve $\displaystyle y = x^2$ is given by $\displaystyle \frac{dy}{dx} = 2x$ .

So at the point $\displaystyle x = 7$ , the slope is $\displaystyle 2\cdot 7 = 14$ .

Equation of tangent: $\displaystyle y = mx + c$ .

You have a coordinate that lies on your tangent $\displaystyle (x, y) = (7, 49)$ and the slope $\displaystyle m = 14$ . Substitute and solve for $\displaystyle c$ , then you can write the equation of your tangent line.

**Quacky** · January 15th 2011, 03:55 PM

How well do you understand the concept behind differentiation?

If $y=x^2,$

$\displaystyle\frac{dy}{dx} = 2x$

$<br /> \displaystyle\frac{dy}{dx}$ is 'the rate of change of y with respect to x'. It is used to tell us the equation of the gradient of the curve at any point.

In the question, $\frac{dy}{dx} = 2x$ . So take any point on your curve. Let's say $(3,9)$ This has x co-ordinate 3.

Substituting this into $\frac{dy}{dx} = 2x$ gives $\frac{dy}{dx} = 6$ . What this means is that at the point $(3,9)$ the gradient of the curve $y=x^2$ is 6.

Also, I don't think that $2x$ is the answer to that question.

We have a point (7,49) and we know that $\frac{dy}{dx} = 2x$ . So the gradient of the tangent is $2\times 7 = 14$ .

So we know that when x = 7, y = 14 and m is 14. Substitute this into y = mx + c to find c and you'll find the equation of the tangent.

**skeeter** · January 15th 2011, 05:30 PM

from first principles ...

$\displaystyle f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

$f(x) = x^2$

$f(x+h) = (x+h)^2$

$\displaystyle f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$

$\displaystyle f'(x) = \lim_{h \to 0} \frac{x^2+2xh+h^2 - x^2}{h}$

$\displaystyle f'(x) = \lim_{h \to 0} \frac{2xh+h^2}{h}<br />$

$\displaystyle f'(x) = \lim_{h \to 0} \frac{h(2x+h)}{h}<br />$

$\displaystyle f'(x) = \lim_{h \to 0} 2x+h = 2x$

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