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A) Step 1: Differentiate ((3x³)/( x)) with respect to x.

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2 parts
a

3x−cot⁡x\frac{3}{x} - \cot x

b

x−12y−8=0x - 12y - 8 = 0

Step-by-step solution

3 steps

Part a

Step 1: Differentiate ln⁡(3x3sin⁡x)\ln\left(\frac{3x^3}{\sin x}\right) with respect to xx.

First, simplify the expression using logarithm properties:

ln⁡(3x3sin⁡x)=ln⁡(3x3)−ln⁡(sin⁡x)\ln\left(\frac{3x^3}{\sin x}\right) = \ln(3x^3) - \ln(\sin x) =ln⁡(3)+ln⁡(x3)−ln⁡(sin⁡x)= \ln(3) + \ln(x^3) - \ln(\sin x) =ln⁡(3)+3ln⁡(x)−ln⁡(sin⁡x)= \ln(3) + 3\ln(x) - \ln(\sin x)

Now, differentiate each term with respect to xx:

ddx(ln⁡(3)+3ln⁡(x)−ln⁡(sin⁡x))\frac{d}{dx}\left(\ln(3) + 3\ln(x) - \ln(\sin x)\right) =ddx(ln⁡(3))+ddx(3ln⁡(x))−ddx(ln⁡(sin⁡x))= \frac{d}{dx}(\ln(3)) + \frac{d}{dx}(3\ln(x)) - \frac{d}{dx}(\ln(\sin x)) =0+3(1x)−(1sin⁡x⋅cos⁡x)= 0 + 3\left(\frac{1}{x}\right) - \left(\frac{1}{\sin x} \cdot \cos x\right) =3x−cos⁡xsin⁡x= \frac{3}{x} - \frac{\cos x}{\sin x} =3x−cot⁡x= \frac{3}{x} - \cot x

The derivative is 3x−cot⁡x\boxed{\frac{3}{x} - \cot x}.

Part b

x−12y−8=0x - 12y - 8 = 0

Step 1: Find dydx\frac{dy}{dx} for the parametric curve.

The curve is defined by x=t3x = t^3 and y=t−14t3y = t - \frac{1}{4}t^3.
First, find dxdt\frac{dx}{dt} and dydt\frac{dy}{dt}:

dxdt=ddt(t3)=3t2\frac{dx}{dt} = \frac{d}{dt}(t^3) = 3t^2 dydt=ddt(t−14t3)=1−14(3t2)=1−34t2\frac{dy}{dt} = \frac{d}{dt}\left(t - \frac{1}{4}t^3\right) = 1 - \frac{1}{4}(3t^2) = 1 - \frac{3}{4}t^2

Now, find dydx\frac{dy}{dx} using the chain rule:

dydx=dy/dtdx/dt=1−34t23t2\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{1 - \frac{3}{4}t^2}{3t^2}

  1. Find the gradient of the tangent at t=−1t = -1.

    Substitute t=−1t = -1 into the expression for dydx\frac{dy}{dx}:

    m=1−34(−1)23(−1)2=1−34(1)3(1)=1−343=143=112m = \frac{1 - \frac{3}{4}(-1)^2}{3(-1)^2} = \frac{1 - \frac{3}{4}(1)}{3(1)} = \frac{1 - \frac{3}{4}}{3} = \frac{\frac{1}{4}}{3} = \frac{1}{12}

  2. Find the coordinates (x,y)(x, y) of the point on the curve at t=−1t = -1.

    Substitute t=−1t = -1 into the parametric equations for xx and yy:

    x=(−1)3=−1x = (-1)^3 = -1 y=(−1)−14(−1)3=−1−14(−1)=−1+14=−44+14=−34y = (-1) - \frac{1}{4}(-1)^3 = -1 - \frac{1}{4}(-1) = -1 + \frac{1}{4} = -\frac{4}{4} + \frac{1}{4} = -\frac{3}{4}

    The point of tangency is (−1,−34)\left(-1, -\frac{3}{4}\right).

  3. Write the equation of the tangent line.

    Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1):

    y−(−34)=112(x−(−1))y - \left(-\frac{3}{4}\right) = \frac{1}{12}(x - (-1)) y+34=112(x+1)y + \frac{3}{4} = \frac{1}{12}(x + 1)

    To eliminate fractions, multiply the entire equation by 12:

    12(y+34)=12(112(x+1))12\left(y + \frac{3}{4}\right) = 12\left(\frac{1}{12}(x + 1)\right) 12y+9=x+112y + 9 = x + 1

    Rearrange the equation into the form Ax+By+C=0Ax + By + C = 0:

    x−12y+1−9=0x - 12y + 1 - 9 = 0 x−12y−8=0x - 12y - 8 = 0

    The equation of the tangent to the curve is x−12y−8=0\boxed{x - 12y - 8 = 0}.

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