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

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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