Find the derivative of the following exponential functions: Write the solution of each number

Question

2.y = x²e^-x

ans. y' = xe^-x(2 - x)

3.f(z) = e²(sin z + cos z)

ans. f'(z)= 2e^z cos z

Answer 1

Kevin,

Problem number 2 uses the product rule and the chain rule. You must also use the formula that

d (e^f(x)) = e^f(x)(f'(x))

dx

So. y = x²e^-x

y' = x² ( e^-x (-1)) + e^-x(2x)

Simplifying

y' = - x² e^-x + 2x e^-x

^{Then factoring}

y' = x e^-x (- x + 2) = x e^-x (2 - x )

Number 3

I believe the problem was typed in with an error. The only way to get the solution given is if the x² is really an x^z.

So that makes it as follows:

f(z) = e^z(sin z + cos z)

It is easiest to distribute the e^z before you do the derivative.

So. f(z) = e^z sin z + e^z cos z Now apply the product rule to each term.

f' (z) = (e^z cos z + e^z sin z) + ( e^z (-sin z) + e^z cos z )

Collecting terms gives

f'(z)= 2e^z cos z

I hope this helps.

Answer 2

For both of these problems, simply use the function multiplication rule:

(uv)' = u'v + uv'

AND the function composition (chain) rule.

2. y=x²e^-x

y' = 2xe^-x + (-e^-xx²) = 2xe^-x - x²e^-x = xe^-x(2-x)

None that I used the commutative law of multiplication and the distributive law above.

3. f(z) = e^z(sin(z) + cos(z))

Note: this must be e^z not e² for the indicated answer to be true.

f'(z) = [e^z(sin(z) + cos(z))]' = [e^zsin(z)]' + [e^zcos(z)]'

= [ e^zsin(z) + e^zcos(z) ] + [ e^zcos(z) - e^zsin(z) ]

= 2e^zcos(z)

NOTE that two of the above terms cancel out.