ShaheenNazar
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If y = e^x cosx, find d^2y/dx^2. If the calculations are done step by step, it would be great!
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Confused with this sum
- Thread starter ShaheenNazar
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Your subject-line says you are "confused with this sum". I do not see any indication of "sum"If y = e^x cosx, find d^2y/dx^2. If the calculations are done step by step, it would be great!
For the problem stated, you should know that:
\(\displaystyle \frac{d}{dx}e^x = e^x\)
continue...
HallsofIvy
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Subhotosh Kahn, in countries heavily influenced by Britain it is not unusual to use the term "sum" to mean any mathematics problem.
Shaheen Nazar, are you taking a Calculus class? This is a Calculus problem and before you could be expected to do it, you should have learned a number of things:
1) that the derivative of \(\displaystyle e^x\) is \(\displaystyle e^x\).
2) that the derivative of cos(x) is -sin(x).
3) that the derivative of sin(x) is cos(x).
3) the "product rule- \(\displaystyle \frac{d(fg)}{dx}= \frac{df}{dx}g+ f\frac{dg}{dx}\).
4) that \(\displaystyle \frac{d^2f}{dx}\) means that you differentiate twice.
Shaheen Nazar, are you taking a Calculus class? This is a Calculus problem and before you could be expected to do it, you should have learned a number of things:
1) that the derivative of \(\displaystyle e^x\) is \(\displaystyle e^x\).
2) that the derivative of cos(x) is -sin(x).
3) that the derivative of sin(x) is cos(x).
3) the "product rule- \(\displaystyle \frac{d(fg)}{dx}= \frac{df}{dx}g+ f\frac{dg}{dx}\).
4) that \(\displaystyle \frac{d^2f}{dx}\) means that you differentiate twice.