# Find the slope of the tangent line to the curve defined by $7x^4 - 8xy - 6y^3 = 322$ at the point $(2, -3)$

Find the slope of the tangent line to the curve defined by

$$7x^4 - 8xy - 6y^3 = 322$$ at the pont $$(2, -3)$$

I'm having a tough time using implicit differentiation and chain rule with all the forms of y in the problem, help is greatly appreciated

• So it appears that you have tried implicit differentiation. What was the form that you got out of it? Is there a specific term that is puzzling to you? Feb 22 '19 at 11:31
• @MattiP. I can't get a final form out of implicit differentiation because I get lost with all the forms of y involving the chain rule Feb 22 '19 at 11:45

Hint: $$\frac{d}{dx} xy = y+x \frac{dy}{dx}$$

If the question does not require you to find the simplest form of $$\frac{dy}{dx}$$ , then just simply put $$(x,y)=(2,-3)$$ after taking the first derivative on both side, after that you will find out what does the slope is.

Key takeaway is that for a function $$y=y(x)$$ then, upon considering

$$z(x) = xy$$

One finds $$z'(x)$$ by first the product rule $$\frac{d}{dx} a(x) b(x) = a \frac{d b}{dx}+ \frac{d a}{dx}b$$ and via 'implicitly differentiating $$y$$ where $$\frac{d}{dx}f(y) = \frac{d}{dy}f(y) \times \frac{dy}{dx}$$ Thus, in my example above, one has \begin{align} \frac{d}{dx}xy &= \frac{dx}{dx}y+x \frac{dy}{dx} \\ &= y+x \frac{dy}{dx} \end{align}

Applying this to the second term in your example above yields \begin{align} \frac{d}{dx}(-8xy) &=\frac{d}{dx}(-8)xy-(8)\frac{d}{dx}(x)y-(8)x\frac{d}{dx}(y) \\ &=-8\frac{d}{dx}(x)y-8x\frac{d}{dx}(y) \\ &= -(8)\frac{dx}{dx}y-8x\frac{dy}{dx} \\ &= -8y-8x\frac{dy}{dx} \\ &= -8y-8xy' \\ \end{align}

Now, clearly, the trickest part of your problem lies in the non-lienar part $$y^3$$ but the application of the above results, along with the 'power rule' (see below) you are aware of already should give you enough momemtum to go forward.

Power Rule: For $$y=y(x)=x^n, \frac{dy}{dx} = n x ^{n-1}$$

It is clear the term in $$y^3$$ is a stumbling block.

Let me walk you through it, using some of the concepts above. Set $$z(x) = 6y^3 = 6y(x)^3$$

Now, \begin{align} \frac{d}{dx} 6y^3&= 6\frac{d}{dy} \times \frac{dy}{dx} y^3 \\ &= 6\frac{d}{dy} y^3 \times \frac{dy}{dx} \\ &= 18y'y^2 \end{align}

Hence

$$28x^3 - 8y - 8xy'-18y'y^2 = 0$$

Or, alternatively $$y' = \frac{28 x^3-8y}{8x+18y^2}$$

Now, insert the points $$(x, y)=(2, -3)$$ and see if this agrees with the answer required.

Further Edit

The Chain rule states For $$z = z(y(x))$$; $$\frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}$$

Here, your y' is indeterminate, ie, this is the form we include in the above 'graident function'. The other part can be calculated explicitly, viz $$z=y^2 = y(x)^3 \, \implies \,\frac{dz}{dx} = \frac{d}{dy}(y^3) \frac{dy}{dx} = 3y^2 \frac{dy}{dx} = 3y^2y'$$

Does this help?

• @SamuelJacobi Good start! for the last bit, inore the factor of $6$, it remains. Now, remember $y=y(x)$ so you must have a derivative of $y$ wrt $x$ as per the 'implicit' rule. Now combine this with the 'power rule' and you should end up with $$6 \times \frac{dy}{dx} \times \text{something in y after power rule}$$ question is, can you find out what the 'something' is?
– user284001
Feb 22 '19 at 12:01
• So far I have 28x^3 - 8y - 8xy' - d/dx(6y^3) = 0 I can't figure out how to apply the chain rule to d/dx(6y^3) Feb 22 '19 at 12:01
• I got 28x^3 - 8y - 8xy' - 18y^2 - 6y' = 0 Feb 22 '19 at 12:04
• Other than getting y', the only other thing I can think of getting is another 18y^2 using your directions Feb 22 '19 at 12:18
• @SamuelJacobi So close Samuel! The reamining two terms you have should be all multiplied together, as $$6\frac{d}{dx}y^3 = 6 \times y' \times 3y^2$$
– user284001
Feb 22 '19 at 12:18