James10492
Junior Member
- Joined
- May 17, 2020
- Messages
- 50
ah, that would be why. Thank youIf [imath]y=c^x~\&~c\in\Re^+[/imath] then [imath]y'=c^x\log(c)[/imath]
The handwriting is a bit light. To make sure I understand the question correctly. You were asked to find the tangent line of the functionHi 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?
Apologies I was writing in pencil, it showed up fine on my screen.Your writing is very light. I tried following your work and it seemed fine.
Exactly that yes.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]?
I agree with your answer.Exactly that yes.
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)If [imath]f(x)=2x+4^{-x}[/imath] then [imath]f^{\prime}(x)=2-\left(4^{-x}\right)\log(4)[/imath]
What you say is exactly what pka said; he is using "log" to mean the same thing as your "ln".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)
Haha. Here is the question in my book (no 39)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.
What's the name and edition of the textbook or ISBN, Chapter and Section?I might be able to see their written solution.