If

f(x) = 5ex cos x, find f '(x) and f ''(x).

To find the derivative of f(x), denoted as f '(x), we can use the product rule and the chain rule.

The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is:

(uv)'(x) = u'(x)v(x) + u(x)v'(x)

In our case, u(x) = 5ex and v(x) = cos x. So, applying the product rule, we have:

f '(x) = (5ex)'(cos x) + (5ex)(cos x)'

To find the derivative of 5ex, we can use the chain rule. The chain rule states that if we have a composite function g(f(x)), then its derivative is:

(g(f(x)))' = g'(f(x)) * f'(x)

In this case, g(u) = 5e^u. So the derivative of 5ex is:

(5ex)' = (5e^u)' * (ex)' = 5e^u * ex = 5e^x

Similarly, to find the derivative of cos x, we differentiate it as a single function:

(cos x)' = -sin x

Now we can substitute the derivatives into f '(x):

f '(x) = (5e^x)(cos x) + (5ex)(-sin x)

Simplifying further, we get:

f '(x) = 5e^x cos x - 5ex sin x

To find the second derivative, denoted as f ''(x), we differentiate f '(x) with respect to x:

f ''(x) = (5e^x cos x - 5ex sin x)'

Again, using the product rule and the chain rule, we have:

f ''(x) = (5e^x cos x)' - (5ex sin x)'

For the first term, we differentiate 5ex cos x using the product rule:

(5e^x cos x)' = (5e^x)'(cos x) + (5e^x)(cos x)'

Applying the chain rule, we find:

(5e^x cos x)' = (5e^x)(cos x) - (5e^x)(sin x)

Simplifying further, we get:

(5e^x cos x)' = 5e^x (cos x - sin x)

For the second term, we differentiate 5ex sin x as:

(5ex sin x)' = (5ex)'(sin x) + (5ex)(sin x)'

Again, using the chain rule, we have:

(5ex sin x)' = (5e^x)(sin x) + (5ex)(cos x)

Now we can substitute the derivatives into f ''(x):

f ''(x) = 5e^x (cos x - sin x) - (5e^x)(sin x + cos x)

Simplifying further, we get:

f ''(x) = 5e^x (cos x - sin x - sin x - cos x)

Finally, combining like terms, we find:

f ''(x) = -10e^x sin x

To find the first derivative, f'(x), of the given function f(x) = 5ex cos x, we can use the product rule and chain rule.

1. Apply the product rule: The derivative of a product of two functions u(x) and v(x) is given by (u'v + uv').

Let u(x) = 5ex and v(x) = cos x.
First, find u'(x) and v'(x):
- The derivative of u(x) = 5ex is u'(x) = 5ex.
- The derivative of v(x) = cos x is v'(x) = -sin x.

Using the product rule formula, we have:
f'(x) = u'v + uv'
= (5ex)(cos x) + (5ex)(-sin x)
= 5ex cos x - 5ex sin x.

Therefore, f'(x) = 5ex cos x - 5ex sin x.

2. To find the second derivative, f''(x), take the derivative of f'(x):

First, find the derivative of f'(x) = 5ex cos x - 5ex sin x.
- The derivative of 5ex cos x is (5ex)' cos x + 5ex(cos x)'
= 5ex cos x - 5ex sin x.
- The derivative of -5ex sin x is (-5ex sin x)' + (-5ex)(sin x)'
= -5ex sin x - 5ex cos x.

Thus, f''(x) = 5ex cos x - 5ex sin x - 5ex sin x - 5ex cos x
= -10ex sin x.

Therefore, f''(x) = -10ex sin x.

ever heard of the product rule?

f = 5e^x cosx
f' = 5e^x cosx - 5e^x sinx
= 5e^x (cosx - sinx)
f" = 5e^x (cosx - sinx) + 5e^x(-sinx-cosx)
= -10e^x sinx

5e^xcosx-5^xsinx