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y=(-2/3)^x

Sep 4, 2024

y=(-2/3)^x

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Solution by Steps

step 1

Differentiate

y = \left(-\frac{2}{3}\right)^x

with respect to

x

step 2

The derivative is given by

y'(x) = -\left(-\frac{2}{3}\right)^x \log\left(\frac{3}{2}\right) + i \pi \left(-\frac{2}{3}\right)^x

step 3

Simplifying, we have

y'(x) = i \left(-\frac{2}{3}\right)^x \left(\pi + i \log\left(\frac{3}{2}\right)\right)

step 4

Setting the derivative equal to zero gives

y'(x) + \left(-\frac{2}{3}\right)^x \left(\log\left(\frac{3}{2}\right) - i \pi\right) = 0

step 5

Thus,

y'(x) = -\left(-\frac{2}{3}\right)^x \left(\log(3) - \log(2)\right) + i \pi \left(-\frac{2}{3}\right)^x

step 6

To find the slope of the tangent line at

(1, 1)

, we evaluate

y'(1)

step 7

However, it is determined that no tangent line exists at this point

Answer

No tangent line exists at the point

(1, 1)

Key Concept

Implicit differentiation and the existence of tangent lines

Explanation

The derivative indicates the slope of the tangent line, but in this case, it shows that no tangent line exists at the specified point.