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The gradient is dydx=kekx\frac{dy}{dx} = ke^{kx}dxdy=kekx.
Differentiate to get dydx=18e3x\frac{dy}{dx} = 18e^{3x}dxdy=18e3x and substitute.
y′=18e3xy' = 18e^{3x}y′=18e3x, apply chain rule for differentiation.
Differentiate, evaluate at ln(2)\ln(2)ln(2), use point-slope form.
Point is (ln(2),6)(\ln(2), 6)(ln(2),6) since 3eln(2)=3⋅2=63e^{\ln(2)} = 3 \cdot 2 = 63eln(2)=3⋅2=6.
y=6x−6ln(2)y = 6x - 6\ln(2)y=6x−6ln(2) after expansion and simplification.
It's the gradient of a tangent at a specific point.
Estimates depend heavily on the curve's quality.
Differentiate the function with respect to xxx.
Differentiate, substitute xxx, and apply point-slope form.
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