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 March 8th, 2010, 05:29 PM #1 Member   Joined: Oct 2009 Posts: 59 Thanks: 0 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: $y= -\frac{1}{f#39;(x_o)}(x - x_o) + y_o$ $y= -8(x - 16) + 6$ $y= -8x + 134$ Help with finding the gradient please.
 March 8th, 2010, 05:54 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,968 Thanks: 1152 Math Focus: Elementary mathematics and beyond 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.
 March 9th, 2010, 05:54 PM #3 Member   Joined: Oct 2009 Posts: 59 Thanks: 0 Re: Derivatives Thanks.

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