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 Tartarus March 8th, 2010 05:29 PM

Derivatives

Could someone, kindly, check these solutions?

$\text {1.) Find the derivative (that is, from first principles) of the function}$ $f(x)= x^3 + 2$ $\text {(x being real) at x.}$
$\text {Hence find the equation of the normal to the curve}$ $y= x^3 + 2$ $\text {at the point (2,10)}$

Eqn. of the normal: $y= -\frac{1}{f#39;(x_o)}(x - x_o) + y_o,$

where $y_o= f(x_o) = f(2) = 10.$

$f'(x) = 3x^2$

$y= -\frac{1}{3\cdot2^2}(x - 2) + 10$

$y= \frac{1}{12}x + \frac{61}{6}$

$\text {2.) Find the derivative (that is, from first principles) of the function}$ $f : x \rightarrow \sqrt{x} + 2\, (x > 0)$ $\text {at x. What is the gradient of the curve}$ $y= f \rightarrow \sqrt{x} + 2$ $\text {at x= 16? Find the equations of the tangent and the normal to the curve}$ $y= f \rightarrow \sqrt{x} + 2$ $\text {at the point (16, 6).}$

Eqn. of the tangent: $y= f#39;(x_o)(x - x_o) + y_o,$

where $y_o= f(x_o) = f(16) = 6$

$f'(x) = \frac{1}{2\sqrt{x}}$

$y= \frac{1}{2\sqrt{16}}(x - 16) + 6$

$y= \frac{1}{8}x + 4$

Eqn. of the normal: $y= -\frac{1}{f#39;(x_o)}(x - x_o) + y_o$

$y= -8(x - 16) + 6$

$y= -8x + 134$

 greg1313 March 8th, 2010 05:54 PM

Re: Derivatives

For 1) $y\,=\,-\frac{x}{12}\,+\,\frac{61}{6}$

For 2) $\text{Gradient of a curve at a point is the value of the derivative evaluated at that point: }f'(16)\,=\,\frac{1}{8}$

Everything else looks correct.

 Tartarus March 9th, 2010 05:54 PM

Re: Derivatives

Thanks.

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