Derivative implicit differentiation problem

In summary, the student is trying to solve a homework equation but is having difficulty. He is getting confused about the rules for differentiating and distributing logs and is not sure how to find dy/dx.
  • #1
laylas
2
0

Homework Statement



Consider the curve satisfying the equation 2(x+1)^(tanx)=(y^2)cosx+y and find dy/dx

Homework Equations


(tanx)'=sec^2x
(lnx)'=1/x

The Attempt at a Solution



I've tried taking the natural log of both sides and then taking d/dx of both sides but something seems to go wrong with my algebra each time.. that or I'm getting confused about my natural log / implicit differentiation rules. I'm not actually sure if I'm approaching the problem correctly but here's one of the many attempts I've made:

ln2 + tanxln(x+1) = 2lny + lncosx + lny(??)
1/2 + sec^2xln(x+1) + tanx(1/x+1) = 2y'/y + sin(x)/cos(x) + y'/y
something should already be wrong so i won't post the rest of the work but i know the answer is supposed to be dy/dx = [2(x+1)^(tanx)[sec^2xln(x+1)+(tanx)/(x+1)]+(y^2)(sinx)]/ all over (2ycosx+1). Need help getting there, thanks!
 
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  • #2
It looks like you tried to distribute the natural log over addition, which doesn't work. The left hand side is fine, but the right hand side should be
[tex]\ln (y^2 \cos(x) + y)[/tex]
Or I suppose you could divide out the y from the original expression, but that won't really help, seeing as you'd just have to combine the two fractions you get after differentiating.
 
  • #3
The hard part for you seems to be differentiating (x+1)^tan(x). f(x)^g(x)=e^(ln(f(x))*g(x)) since f(x)=e^(ln(f(x)). Try differentiating it in that form.
 
  • #4
Once I have the right side down to 1/y^2(cosx+y) [?] how do I find dy/dx?
 
  • #5
laylas said:
Once I have the right side down to 1/y^2(cosx+y) [?] how do I find dy/dx?

If you have the right side down to " 1/y^2(cosx+y) " ... Where did dy/dx go, and what is in the numerator & what in the denominator?

By the way: 1/y^2(cosx+y) literally is the same as (cosx+y)/y2 . It helps to use enough parentheses to say what you mean
 

1. What is implicit differentiation?

Implicit differentiation is a method used to find the derivative of a function that is not explicitly written in terms of the independent variable. This is often used when a function is written in the form of an equation where both the dependent and independent variables are present.

2. How is implicit differentiation different from explicit differentiation?

Explicit differentiation is used to find the derivative of a function that is written explicitly in terms of the independent variable, while implicit differentiation is used for functions that are not written explicitly in terms of the independent variable.

3. What are the steps to solving a derivative implicit differentiation problem?

The general steps to solving a derivative implicit differentiation problem are as follows:
1. Differentiate both sides of the equation with respect to the independent variable.
2. Use the chain rule if necessary to simplify the expression.
3. Solve for the derivative by isolating the derivative term on one side of the equation.
4. Substitute any known values to find the numerical value of the derivative.

4. When is implicit differentiation used in real-life applications?

Implicit differentiation is used in real-life applications when the relationship between variables cannot be explicitly expressed. This can include situations in physics, economics, and engineering, such as finding the rate of change in a system where multiple variables are present.

5. Are there any limitations or restrictions when using implicit differentiation?

Yes, there are some limitations when using implicit differentiation. It can only be used when the function is differentiable, and it can become more complicated when dealing with higher-order derivatives. Additionally, it may not be as intuitive as explicit differentiation for some individuals.

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