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When trying to find the first partial derivatives for the function $\psi(x,t)=ae^{-(bx+ct)^2}$,

I am getting the following answers:

$$\frac{ \partial \psi}{\partial x}=2bae^{-(bx+ct)^2}$$

and

$$\frac{ \partial \psi}{\partial t}=2cae^{-(bx+ct)^2} $$

This does not appear to be correct. Could someone please explain what I am doing wrong?

Siong Thye Goh
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LearnIcon
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1 Answers1

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Use chain rule,\begin{align}\frac{ \partial \psi}{\partial x}&=ae^{-(bx+ct)^2}\frac{\partial}{\partial x}(-(bx+ct)^2)\\&=ae^{-(bx+ct)^2}(-2(bx+ct))\frac{\partial}{\partial x}(bx+ct)\\&=ae^{-(bx+ct)^2}(-2b(bx+ct)) \end{align}

Similarly for $\frac{ \partial \psi}{\partial t}.$

Siong Thye Goh
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  • Thank you for your response and clarification. I can see where I went wrong now. If I need to take the second partial derivative and so on, does that mean I need to use product rule then chain rule? – LearnIcon Aug 26 '17 at 06:47
  • Yes. Also note that $\frac{\partial \psi}{\partial x} = \psi (-2b(bx+ct))$ to cut down repeated computation. – Siong Thye Goh Aug 27 '17 at 02:37