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Differentiation of Exponential Functions - Question

How do I go about differentiating xe^{x^2}?

I started by letting u = x and v = e^{x^2} , so du/dx = 1 but I don't know how to find dv/dx.

Any help would be greatly appreciated!

Thanks!

Reply 1

ghostwalker

Original post by grace15

How do I go about differentiating xe^{x^2}?

I started by letting u = x and v = e^{x^2} , so du/dx = 1 but I don't know how to find dv/dx.

Any help would be greatly appreciated!

Thanks!

Use the chain rule on

e^{x^2}

Let

w=x^2

(edited 6 years ago)

Reply 2

grace15

Original post by ghostwalker

Use the chain rule on

Unparseable latex formula:

e^x^2

So if I let y= (e^x)²then dy/dx = 2 (e^x )(e^x ) = 2e^{x^2}?

Reply 3

username2028267

So, you have to do x' * (e^(x^2)) + x*((e^(x^2))'
Its the formula u'v + uv'

Reply 4

ghostwalker

Original post by grace15

So if I let y= (e^x)²then dy/dx = 2 (e^x )(e^x ) = 2e^{x^2}?

I edited my previous post to suggest the substitution.

NOTE:

e^{x^2}\not=(e^x)^2

Edit:

(e^x)^2 =e^x\times e^x = e^{2x}

(edited 6 years ago)

Reply 5

grace15

So if I let v=e^w and let w = x² then dy/dx of e^{x^2} is 2xe^{x^2} ?

Then would I use the product rule to get dy/dx = x(2xe^{x^2} ) + e^{x^2} ?

Which would simplify to e^{x^2} (2x² + 1) ?

Is this correct?

Yes.

Original post by grace15