Find the slope of the tangent line to the graph of y=(3-8x)^1/3 at the point (-3,3).

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The slope of the tangent line to the graph of y=f(x) at any point (x, f(x)) is equal to the value of the first derivative of the graph at that point.

The first derivative of y (also called the gradient function of the curve) is then found as follows:

Given that `y = (3-8x)^{\frac{1}{3}}` , then we let `u = 3 - 8x` , so that `y = u^{\frac{1}{3}}` .

Therefore, if `y = u^{\frac{1}{3}}` , then `\frac{dy}{du} = \frac{1}{3} u^{\frac{1}{3} - \frac{3}{3}} = \frac{1}{3}u^{\frac{-2}{3}} = \frac{1}{3}(3-8x)^{\frac{-2}{3}}` (use the power rule of differentiation and substitute for `u` in the answer).

Further, if `u = 3-8x` , then `\frac{du}{dx} = -8` .

From the chain rule of differentiation, if `y = f(u)` and `u = g(x)` then `\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}` . Therefore, `\frac{dy}{dx}` in the given problem is `\frac{1}{3}(3-8x)^{\frac{-2}{3}} \times -8 = \frac{-8}{3}(3-8x)^{\frac{-2}{3}}` .

The slope of the tangent line at the point (-3, 3) is found by substituting `x=-3` into the gradient function obtained above:

`\frac{-8}{3} (3-8(-3))^{\frac{-2}{3}} = \frac{-8}{3}(27)^{\frac{-2}{3}} = \frac{-8}{3} \times \frac{1}{9} = \frac{-8}{27}` .

Thus the slope of the tangent line at the point (-3, 3) is `\frac{-8}{27}` .

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