Answer :
The question is:
Find the derivatives of the following functions
(1) y = x^(11/5)
(2) f(x) = 1/(x^16)
(3) y = x²(9x + 3)
(4) f(z) = -1/(z^3.9)
(5) y = (√x)^(1/14)
(6) f(x) = √(1/x³)
(7) y = 7t² + 10e^t
(8) g(x) = (2x³ + 1)^5
(9) w = (t^5 + 1)^92
(10) y = √(s^11 + 2)
(11) y = 2e^(4t + 3)
(12) y = ln(2t + 3)
(13) f(x) = ln(e^x + 5)
Step-by-step explanation:
If y = x^n, the derivative of y with respect to x is written as dy/dx or y', and it is given as
y' = nx^(n - 1).
If y = f(u), and u = u(x),
y' = f'(u) × u'(x)
If y = e^x, y' = e^x
If y = ln(u) and u = u(x)
y' = (1/u) × du/dx
Now, let us solve the problems given.
Answer:
(1) y = x^(11/5)
y' = (11/5)x^(11/5 - 1)
= (11/5)x^(6/5)
(2) f(x) = 1/(x^16)
This can be written as
f(x) = x^(-16)
f'(x) = -16x^(-16 - 1)
= -16x^(-17)
(3) y = x²(9x + 3)
y' = 2x(9x + 3) + x²(9)
= 18x² + 6x + 9x²
= 27x² + 6x
(4) f(z) = -1/(z^3.9)
This can be written as
f(z) = z^(-3.9)
f'(z) = -3.9z^(-3.9 - 1)
= -3.9z^(-4.9)
(5) y = (√x)^(1/14)
This can be written as
y = x^(1/28)
y' = (1/28)x^(1/28 - 1)
= (1/28) x^(-27/28)
(6) f(x) = √(1/x³)
This can be written as
f(x) = x^(-3/2)
f'(x) = (-3/2)x^(-3/2 - 1)
= (-3/2)x^(-5/2)
(7) y = 7t² + 10e^t
y' = 14t + 10e^t
(8) g(x) = (2x³ + 1)^5
g'(x) = 6x² × 5(2x³ + 1)^4
= 30x²(2x³ + 1)^4
(9) w = (t^5 + 1)^92
w' = 5t × 92(t^5 + 1)^91
= 460t(t^5 + 1)^91
(10) y = √(s^11 + 2)
This can be written as
y = (s^11 + 2)^(1/2)
y = 11s × (1/2)(s^11 + 2)^(-1/2)
= (11s/2)(s^11 + 2)^(-1/2)
(11) y = 2e^(4t + 3)
y' = 4 × 2e^(4t + 3)
= 8e^(4t + 3)
(12) y = ln(2t + 3)
y' = 2/(2t + 3)
(13) f(x) = ln(e^x + 5)
f'(x) = e^(x)/(e^x + 5)
The derivatives of the following functions calculated by using the derivation identities.
1-
[tex]y = x^{(11/5)}[/tex]
[tex]y' = \dfrac{11}{5} x^{(11/5 - 1)}[/tex]
[tex]y'=\dfrac{11}{5} x^{(6/5)}[/tex]
2-
[tex]f(x) =\dfrac{1}{ x^{16}}[/tex]
[tex]f(x)= x^{-16}[/tex]
[tex]f'(x)=-16x^{(-16-1)}[/tex]
[tex]f'(x)=-16x^{-17}[/tex]
3-
[tex]y = x^2(9x + 3)[/tex]
[tex]y' = 2x(9x + 3) + x^2(9)[/tex]
[tex]y'= 18x^2 + 6x + 9x^2[/tex]
[tex]y'= 27x^2 + 6x[/tex]
4-
[tex]f(z) =\dfrac{-1}{z^{3.9}}[/tex]
[tex]f(z) ={z^{-3.9}}[/tex]
[tex]f'(z) =-3.9{z^{-3.9-1}}[/tex]
[tex]f'(z) =-3.9{z^{-4.9}[/tex]
5-
[tex]y = (\sqrt{x} )^{(1/14)}[/tex]
[tex]y = x^{(1/28)}[/tex]
[tex]y' = \dfrac{1}{28} x^{(1/28-1)}[/tex]
[tex]y' = \dfrac{1}{28} x^{(-27/28)}[/tex]
6-
[tex]f(x) = \sqrt{(\dfrac{1}{x^3} )}[/tex]
[tex]f(x) =x^{(-3/2)}[/tex]
[tex]f'(x) =-\dfrac{3}{2} x^{(-3/2-1)}[/tex]
[tex]f'(x) =-\dfrac{3}{2} x^{(-5/2)}[/tex]
7-
[tex]y = 7t^2+ 10e^t[/tex]
[tex]y' = 14t + 10e^t[/tex]
8-
[tex]g(x) = (2x^3+ 1)^5[/tex]
[tex]g'(x) = 6x^2 \times 5(2x^3 + 1)^4[/tex]
[tex]g'(x)= 30x^2(2x^3 + 1)^4[/tex]
9-
[tex]w = (t^5 + 1)^{92}[/tex]
[tex]w' = 5t \times92(t^5 + 1)^{91}[/tex]
[tex]w'= 460t(t^5 + 1)^{91}[/tex]
10-
[tex]y = \sqrt{(s^{11} + 2)}[/tex]
[tex]y = (s^{11} + 2)^{(1/2)}[/tex]
[tex]y = 11s \times\dfrac{1}{2} (s^{11} + 2)^{(-1/2)}[/tex]
[tex]y'= \dfrac{11s}{2} (s^{11} + 2)^{(-1/2)}[/tex]
11-
[tex]y = 2e^{(4t + 3)}[/tex]
[tex]y' = 4 \times2e^{(4t + 3)}[/tex]
[tex]y'= 8e^(4t + 3)[/tex]
12-
[tex]y = ln(2t + 3)\\y' = \dfrac{2}{(2t + 3)}[/tex]
[tex]f(x) = ln(e^x + 5)f'(x) = e^{\dfrac{x}{(e^x + 5)}[/tex]
Hence, the derivatives of the following functions calculated by using the derivation identities.
For more about the derivation follow the link given below-
https://brainly.com/question/2788760