Finding the equation of a tangent involving a derivative function with natural logarithms

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James10492

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Hi again, having some difficulty with this problem which I have tried but cannot solve. I must be missing something but I cannot decide what.

tangent problem.jpgtangent problem continued.jpg

I cannot see how there is 15ln2. Where did I go wrong?
 
If [imath]y=c^x~\&~c\in\Re^+[/imath] then [imath]y'=c^x\log(c)[/imath]
 
James10492 said:
Hi again, having some difficulty with this problem which I have tried but cannot solve. I must be missing something but I cannot decide what.

View attachment 32087View attachment 32088

I cannot see how there is 15ln2. Where did I go wrong?
Click to expand...
The handwriting is a bit light. To make sure I understand the question correctly. You were asked to find the tangent line of the function
[imath]f(x)=2x+4^{-x}[/imath] at the point [imath]x=-1[/imath]?
 
Your writing is very light. I tried following your work and it seemed fine.
 
If [imath]f(x)=2x+4^{-x}[/imath] then [imath]f^{\prime}(x)=2-\left(4^{-x}\right)\log(4)[/imath]
 
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pka said:
If [imath]f(x)=2x+4^{-x}[/imath] then [imath]f^{\prime}(x)=2-\left(4^{-x}\right)\log(4)[/imath]
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please could you show me the proof for that? In my working I get to -ln4 4^-x (using the same method for differentiating a^x)
 
James10492 said:
please could you show me the proof for that? In my working I get to -ln4 4^-x (using the same method for differentiating a^x)
Click to expand...
What you say is exactly what pka said; he is using "log" to mean the same thing as your "ln".

So I agree with your first page of work, where you get y = (2-4ln4)x + 4(1-ln4), though I simplified mine to y = 2(1-4ln2)x + 4(1-2ln2).

On the second page, you appear to be trying to make your answer look like the supposed answer, (15ln2)x + 2y + 15ln2 - 4 = 0.

Are you sure that answer is correct? Here are the graphs of the curve and the two claimed tangent lines:

1649788355740.png​

Your line is in red, and is correct. Theirs is in green, and clearly is not tangent.
 
Say that [imath]a\in\Re^+[/imath] and [imath]y=a^{-x}.[/imath]
In order to use implicit differentiation, [imath]\log(y)=-x\log(a)[/imath]
Note that [imath]\log(a)[/imath] is a constant. The derivative is:
[imath]\dfrac{y^{\prime}}{y}=-\log(a)[/imath] or [imath]y^{\prime}=-y\log(a)=-a^{-x}\log(a)[/imath]

[imath][/imath][imath][/imath][imath][/imath]
 
Dr.Peterson said:
What you say is exactly what pka said; he is using "log" to mean the same thing as your "ln".

So I agree with your first page of work, where you get y = (2-4ln4)x + 4(1-ln4), though I simplified mine to y = 2(1-4ln2)x + 4(1-2ln2).

On the second page, you appear to be trying to make your answer look like the supposed answer, (15ln2)x + 2y + 15ln2 - 4 = 0.

Are you sure that answer is correct? Here are the graphs of the curve and the two claimed tangent lines:


Your line is in red, and is correct. Theirs is in green, and clearly is not tangent.
Click to expand...
Haha. Here is the question in my book (no 39)
 
Looking at the exam list you posted #39 is [imath]f(x)=2x+4^{-x}[/imath].
You are asked for the equation of the tangent to the graph of [imath]f[/imath] at the point where [imath]x=-1.[/imath].
[imath]f(x)=2x+4^{-x}\\f^{\prime}(x)=2-4^{-x}\log(4)\\f(-1)=-2+4=2\\f^{\prime}(-1)=2-4^{1}\log(4)=2-8\log(2)[/imath]
So we want the tangent to [imath]f(x)=2x+4^{-x}[/imath] at the point [imath](x_0,y_0)=(-1,2)[/imath].
Well the slope there is [imath]f^{\prime}(-1)=m=2-8\log(2)[/imath]
Thus the equation is [imath](y-2)=[2-8\log(2)](x+1)[/imath] SEE HERE
That is [imath](y-y_0)=m(x-x_0)[/imath] and is not the given answer.
[imath][/imath]
 
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