Calculus 10th Edition

$y=3x - 2$
You begin by finding the derivative of the given function: $x^{3}$ You can begin by using the limiting process: $\lim\limits_{\Delta x \to 0} \frac {f(\Delta x+x)-f(x)} {\Delta x}$ Substituting the given function in: $\lim\limits_{\Delta x \to 0} \frac {(x+\Delta x)^{3}-x^{3}} {\Delta x}$ This can now be multiplied out: $\lim\limits_{\Delta x \to 0} \frac {x^{3}+3x^{2}\Delta x + 3x \Delta x^{2} + \Delta x^{3} - x^{3}} {\Delta x}$ Delta x can then be divided out and the x cubes will cancel: $\lim\limits_{\Delta x \to 0} 3x^{2} + 3x \Delta x + \Delta x^{2}$ Applying the limit gives you $3x^{2}$ This must then equal the slope of the given equation because the lines are parallel, putting the equation in slope intercept form yields: $y = 3x +1$ therefore the slope is 3 and 3 must equal the derivative of the function. Dividing both sides by 3 and taking the square root yields Positive and negative 1. Only one is required for the problem so positive 1 will be chosen. $f(1)=1^{3}$, therefore, the tangent line passes through (1, 1). Now the equation is left. $y=3x+b$ $1 = 3(1) +b$ $b=-2$ And the solution can now be written as $y=3x -2$