Question

Find the first partial derivatives of the function. f(x, t) = e−9t cos(πx)

Answer 1

Answer:

$f_{x}(x,t) = -\pi e^{-9t} \sin{(\pi x)}$

$f_{t}(x,t) = -9\cos{(\pi x)} e^{-9t}$

Step-by-step explanation:

We are given the following function:

$f(x,t) = e^{-9t} \cos{(\pi x)}$

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.

$f_{x}(x,t) = e^{-9t} \frac{d}{dx](\cos{(\pi x)}) = -e^{-9t} \sin{(\pi x)} \frac{d}{dx}(\pi x) = -\pi e^{-9t} \sin{(\pi x)}$

Function of t:

Same logic as above, the cosine as treated as a constant.

$f_{t}(x,t) = \cos{(\pi x)} \frac{d}{dt}(e^{-9t}) = \cos{(\pi x)} e^{-9t} \frac{d}{dt}(-9t) = -9\cos{(\pi x)} e^{-9t}$