0 + 2 = .
- f′(x) = ex cos x (cos x − x sin x)
The rule for finding the derivative of y = eu is = eu , where u is a function of x. Here we
will use the Product Rule to find the derivative of the exponent: f′(x) = ex cos x (cos x − x sin x).
- f′(x) = − 3e−3x sin 5x + 5e−3x cos 5x
The rule for finding the derivative of y = eu is = eu , where u is a function of x. Here we
will use the Product Rule to find the derivative: f′(x) = e−3x(−3)sin 5x + e−3x(5 cos 5x), which
we can rearrange to f′(x) = −3e−3x sin 5x + 5e−3x cos 5x.
- f′(x) = πeπx − π
The rule for finding the derivative of y = eu is = eu , where u is a function of x and the
rule for finding the derivative of y = ln u is , where u is a function of x.
We get f′(x) = eπx(π) − πeπx. This can be simplified to f′(x) = πeπx − π. You might have
noticed that ln eπx = πx, which would have made the derivative a little easier.
- f′(x) =
The rule for finding the derivative of y = loga u is , where u is a function of x.
Before we find the derivative, we can use the laws of logarithms to expand the logarithm. We
get f(x) = log 12 x^3 = 3 log 12 x. Now we can find the derivative: f′(x) = .
- f′(x) =
The rule for finding the derivative of y = loga u is , and the rule for finding the
