Find dy/dx at the point (0,1), if y5 = (x + 2)4 + ex ln y-15.

Question

Find dy/dx at the point (0,1), if y⁵ = (x + 2)⁴ + e^x ln y-15.

 

This question has me, I've tried it numerous times yet it doesn't seem to work.

Answer 1

Find dy/dx at the point (0,1), if y5 = (x + 2)4 + ex ln y-15.

 

I assume that this question means to state:

 

Find dy/dx at the point (0,1), if y⁵ = (x +2)⁴ + e^xln(y) -15

 

If, instead, the question had meant to state:

 

Find dy/dx at the point (0,1), if y⁵ = (x +2)⁴ + e^xln(y-15) 

 

then the question would not have any answer because the point (0, 1) would not be located on the function.  (Try plugging "0" in for x, and "1" in for y, to see that this is so).  dy/dx is only defined for points (x, y) that are on the function.

 

Let's take the derivative of both sides (of the correct expression!):

 

5y⁴y' = 4(x+2)³(x+2)' + [e^xx'][ln(y)] +[e^x][y^-1y`] + 0                                                             (1)

 

Next, we simplify.  Note that the derivative of x or any term (x + constant) is always equal to  1.

 

5y⁴y' = 4(x+2)³ + e^xln(y) + e^xy^-1y'                                                                                        (2)

 

Next, we collect the y' terms:

 

y'[5y⁴ - e^xy^-1] = 4(x-2)³ + e^xln(y)                                                                                           (3)

 

y' = [4(x-2)³ + e^xln(y)] / [5y⁴ - e^xy^-1]                                                                                     (4)

 

Next, we plug in our point (0, 1):

 

y' = [4(0-2)³ + e⁰ln(1)] / [5(1)⁴ - e⁰(1)^-1]                                                                                (5)

 

y' = [4(-8) + 1(0)] / [5(1) - 1(-1)]                                                                                             (6)

 

y' = [-32 +0] / [5 +1]                                                                                                               (7)

 

y' = -32/5                                                                                                                                (8)

 

 

 

 

 

Answer 2

In cases where it's not easy or possible to rearrange the equation in a y=f(x) form, you should use implicit differentiation coupled with the chain rule.

To put it simply, for implicit differentiation, you take the derivative of every single term on both sides of the equation as you normally would using the derivative rules but to satisfy the chain rule, you must ultimately relate that derivative to x and you do that by multiplying that derivative with the term dV/dx where V is the variable in the term. If the term contains multiple variables then you must apply the product rule.

Let's look at your question (I'm not sure if the term inside the ln is y or (y-15) so I'll assume the latter given the lack of spaces around the - sign):

 

What are the terms in the equation? 

 

1. y⁵

2. (x+2)⁴

3. e^xln(y-15)

 

So let's derive them separately:

 

1. d/dx(y⁵) = 5y⁴(dy/dx)

2. d/dx[(x+2)⁴] = 4(x+2)³(dx/dx) = 4(x+2)³

3. d/dx[e^xln(y-15)] = e^x(dx/dx)ln(y-15)+e^x[1/(y-15)]*dy/dx = e^xln(y-15)+e^x*(dy/dx)/(y-15)

 

So plug the derivatives of those back into the equation:

 

5y⁴*dy/dx = 4(x+2)³+e^xln(y-15)+e^x*(dy/dx)/(y-15)

 

Normally to find dy/dx you'd solve for dy/dx and isolate on one side of the equation. Since you are told to find its value at the point (0,1), you can instead just plug in x=0 and y=1 into the above equation, simplify, and solve for dy/dx:

 

5(1)⁴*dy/dx = 4(0+2)³+e⁰ln(1-15)+e⁰*(dy/dx)/(1-15)

 

5*dy/dx = 32+ln(-14)-(dy/dx)/14

 

71/14*dy/dx = 32+ln(-14) 

 

The answer is an imaginary number because of the negative natural log but easily done on a calculator.