Differentiation problem

[jonnburton]

Hi, can anyone help me with a differentiation problem I'm having trouble with?

The following is the question and what I've done:

Find the equation of the tangent to the curve with equation \(\displaystyle y=\sqrt{x}e^x\) a the point where x = 4

My working:

\(\displaystyle y= 2e^4\)

\(\displaystyle \frac{dx}{dy} = \frac{e^x}{2\sqrt{x}}\)

Putting the value x = 4 into the derivative gives:

\(\displaystyle \frac{e^4}{4}\)



so, for an equation of the tangent:

\(\displaystyle y - 2e^4 = \frac{e^4}{4}(x-4)\)

\(\displaystyle y - 2e^4 = \frac{xe^4}{4}-e^4\)

\(\displaystyle y = \frac{xe^4}{4}+e^4\)


However, the answer give in the book is different; it is supposed to be:

\(\displaystyle 4y+28e^4 = 9e^4x\)

I would be very grateful for any help!

 

[MarkFL]

You have made an error in finding the derivative. You want to use the product rule. You have the correct point \(\displaystyle (4,2e^4)\), but your slope is wrong because of the aforementioned incorrect derivative.

 

[jonnburton]

Thanks a lot MarkFL.

I seem to have missed out part of the derivative.

I think it should be:

\(\displaystyle \sqrt{x}e^x + \frac{1}{2\sqrt{x}}e^x\) = \(\displaystyle 2x e^x + e^x\) = ^\(\displaystyle e^x(2x +1)\)

therefore when x =4, gradient = \(\displaystyle 9e^4\)

x =4, y = \(\displaystyle 2e^4\)

\(\displaystyle y - 2e^4 = 9e^4 (x - 4) \) = \(\displaystyle y - 2e^4 = 9e^4x - 36e^4\)

=\(\displaystyle y = 9e^4x - 36e^4 + 2 e^4\)

\(\displaystyle y = 9e^4x - 34e^4\)

It is beginning to make more sense now but I can't see why the answer is not the same as the book's answer, given that I've got the gradient and the coordinate values for x and y.

 

[daon2]

Thanks a lot MarkFL.

I seem to have missed out part of the derivative.

I think it should be:

\(\displaystyle \sqrt{x}e^x + \frac{1}{2\sqrt{x}}e^x\) = \(\displaystyle 2x e^x + e^x\) = ^\(\displaystyle e^x(2x +1)\)

Your problem is here. Your derivative is fine but you "simplified" it wrong and i havent a clue what it says or what you did. The slope is 9/4*e^4

therefore when x =4, gradient = \(\displaystyle 9e^4\)

x =4, y = \(\displaystyle 2e^4\)

\(\displaystyle y - 2e^4 = 9e^4 (x - 4) \) = \(\displaystyle y - 2e^4 = 9e^4x - 36e^4\)

=\(\displaystyle y = 9e^4x - 36e^4 + 2 e^4\)

\(\displaystyle y = 9e^4x - 34e^4\)

It is beginning to make more sense now but I can't see why the answer is not the same as the book's answer, given that I've got the gradient and the coordinate values for x and y.

 

[HallsofIvy]

Thanks a lot MarkFL.

I seem to have missed out part of the derivative.

I think it should be:

\(\displaystyle \sqrt{x}e^x + \frac{1}{2\sqrt{x}}e^x\) = \(\displaystyle 2x e^x + e^x\) = ^\(\displaystyle e^x(2x +1)\)

No, you cannot just multiply a function by \(\displaystyle 2\sqrt{x}\) and have the same function! Perhaps you meant to write \(\displaystyle \frac{(2x+ 1)e^x}{2\sqrt{x}}\)

therefore when x =4, gradient = \(\displaystyle 9e^4\)

x =4, y = \(\displaystyle 2e^4\)

\(\displaystyle y - 2e^4 = 9e^4 (x - 4) \) = \(\displaystyle y - 2e^4 = 9e^4x - 36e^4\)

=\(\displaystyle y = 9e^4x - 36e^4 + 2 e^4\)

\(\displaystyle y = 9e^4x - 34e^4\)

It is beginning to make more sense now but I can't see why the answer is not the same as the book's answer, given that I've got the gradient and the coordinate values for x and y.

 

[jonnburton]

OK thanks, HallsofIvy and Daon2; I see that was a stupid mistake... I was working on it just now and it works out fine when differentiated properly!

Thanks a lot for your help.