Find the derivative of the following functions:
a) f(x)=(3x^2-6 )/(x+2)
b) g(x)=sin(e^2 +5x)
c) j(x)=6xe^x
d) Your velocity is v(t)= t^2 = 1 for 0≤ t ≤ 4. Find or estimate the distance traveled during this time.
F'(x) = 3(x^2 +4x +2/(x+2)^2
F'(x) = 5cos(5x + e^2)
F'(x) = 6x(e^x) + e^x (6)
= 6e^x (x +1)
a. F'(x) = 3(x^2 +4x +2/(x+2)^2
b. F'(x) = 5cos(5x + e^2)
c. F'(x) = 6x(e^x) + e^x (6)
= 6e^x (x +1)
Thank you very much!
To find the derivatives of the given functions, I'll explain the steps for each one:
a) f(x) = (3x^2 - 6)/(x + 2)
To find the derivative of this function, we can use the quotient rule. The quotient rule states that for a function u(x)/v(x), the derivative is given by (v(x)u'(x) - u(x)v'(x))/(v(x))^2.
In this case, u(x) = (3x^2 - 6) and v(x) = (x + 2).
Now, differentiate u(x) and v(x) separately:
u'(x) = 6x
v'(x) = 1
Substituting these values into the quotient rule formula, we get:
f'(x) = [(x + 2)(6x) - (3x^2 - 6)(1)]/[(x + 2)^2]
Simplifying further, we get:
f'(x) = (6x^2 + 12x - 3x^2 + 6)/[(x + 2)^2]
f'(x) = (3x^2 + 12x + 6)/[(x + 2)^2]
b) g(x) = sin(e^2 + 5x)
To find the derivative of this function, we need to use the chain rule. The chain rule states that if we have a composite function g(f(x)), the derivative is given by g'(f(x)) * f'(x).
In this case, f(x) = e^2 + 5x and g(x) = sin(x).
Differentiating f(x), we get:
f'(x) = 5
Now, differentiating g(x) with respect to x, we get:
g'(x) = cos(x)
Using the chain rule, we have:
g'(f(x)) * f'(x) = cos(e^2 + 5x) * 5
So, the derivative of g(x) is:
g'(x) = 5cos(e^2 + 5x)
c) j(x) = 6xe^x
To find the derivative of this function, we use the product rule. The product rule states that for two functions u(x) and v(x), the derivative of their product u(x)v(x) is given by u'(x)v(x) + u(x)v'(x).
In this case, u(x) = 6x and v(x) = e^x.
Differentiating u(x), we get:
u'(x) = 6
Differentiating v(x), we get:
v'(x) = e^x
Now, applying the product rule, we have:
j'(x) = (6x)(e^x) + (6)(e^x)
j'(x) = 6xe^x + 6e^x
d) For the last part, we have the velocity function v(t) = t^2 - 1 for 0 ≤ t ≤ 4. To find or estimate the distance traveled during this time, we need to integrate the absolute value of the velocity function over the given time interval.
The distance traveled is given by the integral of |v(t)| dt from t=0 to t=4.
First, let's find the absolute value of v(t):
|v(t)| = |t^2 - 1|
Next, we integrate |v(t)| from t=0 to t=4:
Distance traveled = ∫(|v(t)|)dt from t=0 to t=4
This involves breaking up the integral into two parts based on the behavior of v(t):
For t in the interval [0, 1], v(t) = t^2 - 1 is negative. Therefore, the integral becomes:
∫(-(t^2 - 1))dt from t=0 to t=1
For t in the interval [1, 4], v(t) = t^2 - 1 is positive. Therefore, the integral becomes:
∫(t^2 - 1)dt from t=1 to t=4
Evaluating both integrals will give you the total distance traveled during the given time interval.