We can use the chain rule to find the derivative of h(y):
h'(y) = (1/(yln2)) * (2y/((2+y^2)ln2))
Simplifying this expression, we get:
h'(y) = 2/(y(2+y^2)ln2)
Therefore, the derivative of h(y) is h'(y) = 2/(y(2+y^2)ln2).
What is the derivative of
h(y) = log
y
2 + 4y
?
3 answers
By applying differentiation,
sketch the graph of
k(x) = 20x
3 − 3x
5
.
sketch the graph of
k(x) = 20x
3 − 3x
5
.
To sketch the graph of k(x), we need to find the derivative of k(x) and analyze its behavior:
k(x) = 20x^3 - 3x^-5
k'(x) = 60x^2 + 15x^-6
Setting k'(x) equal to zero to find critical points:
60x^2 + 15x^-6 = 0
Dividing both sides by 15x^2, we get:
4x^-4 + 1 = 0
Solving for x, we get:
x = (1/4)^(1/4) or x = -(1/4)^(1/4)
These are the only critical points since k'(x) is always positive or always negative in between them.
Now, we can analyze the behavior of k(x) and sketch the graph:
- As x approaches negative infinity, k(x) approaches negative infinity.
- As x approaches positive infinity, k(x) approaches positive infinity.
- k(x) is decreasing on the interval (-infinity, (1/4)^(1/4)).
- k(x) is increasing on the interval ((1/4)^(1/4), (1/4)^(1/4)).
- k(x) is decreasing on the interval ((1/4)^(1/4), infinity).
Using this information, we can sketch the graph of k(x) as shown below:
k(x) = 20x^3 - 3x^-5
k'(x) = 60x^2 + 15x^-6
Setting k'(x) equal to zero to find critical points:
60x^2 + 15x^-6 = 0
Dividing both sides by 15x^2, we get:
4x^-4 + 1 = 0
Solving for x, we get:
x = (1/4)^(1/4) or x = -(1/4)^(1/4)
These are the only critical points since k'(x) is always positive or always negative in between them.
Now, we can analyze the behavior of k(x) and sketch the graph:
- As x approaches negative infinity, k(x) approaches negative infinity.
- As x approaches positive infinity, k(x) approaches positive infinity.
- k(x) is decreasing on the interval (-infinity, (1/4)^(1/4)).
- k(x) is increasing on the interval ((1/4)^(1/4), (1/4)^(1/4)).
- k(x) is decreasing on the interval ((1/4)^(1/4), infinity).
Using this information, we can sketch the graph of k(x) as shown below:
