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Sagot :
Answer:
[tex]f_{x}(x,t) = -\pi e^{-9t} \sin{(\pi x)}[/tex]
[tex]f_{t}(x,t) = -9\cos{(\pi x)} e^{-9t}[/tex]
Step-by-step explanation:
We are given the following function:
[tex]f(x,t) = e^{-9t} \cos{(\pi x)}[/tex]
First derivatives:
We find the first derivatives in function of x and of t.
Function of x:
The exponential is only a function of t, so it is treated as a constant.
[tex]f_{x}(x,t) = e^{-9t} \frac{d}{dx](\cos{(\pi x)}) = -e^{-9t} \sin{(\pi x)} \frac{d}{dx}(\pi x) = -\pi e^{-9t} \sin{(\pi x)}[/tex]
Function of t:
Same logic as above, the cosine as treated as a constant.
[tex]f_{t}(x,t) = \cos{(\pi x)} \frac{d}{dt}(e^{-9t}) = \cos{(\pi x)} e^{-9t} \frac{d}{dt}(-9t) = -9\cos{(\pi x)} e^{-9t}[/tex]
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