Question

COSX х Find the derivative of the following functions following instructions. Put all answers in positive exponential form. (Any 4) (4 points each) a) f(x) = + 5* – log(x) – log(2) + 2€. Find derivative using properties of derivatives. b) l(x) = cot(5ex + 1) can be expressed as a composition of two functions I and O, where I is the inner function and O is the outer function. Identify I and O. Write down the chain rule using letters I and O. Apply chain rule to find its derivative. c) g(x) = tan-'(5x)+cot-'(5x). Each of the functions can be expressed as a composition of two functions I and 0, where I is the inner function and O is the outer function. Identify I and O for each of the functions involved in g(x). Write down the chain rule using letters I and O. Apply chain rule and then other properties of derivatives to complete your solution. d) k(x) = ln(2x – 1) +€?* – 13x+4 . Each of the functions can be expressed as a composition of two functions I and 0, where I is the inner function and O is the outer function. Identify I and O for each of the functions involved in k(x). Write down the chain rule using letters I and O. Apply chain rule and then other properties of derivatives to complete your solution.. In (sinx) d) r(x) Find derivative using properties of derivatives. = e5x

Answer 1

a) To find the derivative of f(x) = + 5* – log(x) – log(2) + 2€, we can use the properties of derivatives. Let's break it down step by step:

First, we know that the derivative of a constant term is always zero. So, the derivative of 2€ is zero.

Next, the derivative of 5* is also zero, as it's a constant term.

Now, let's differentiate the logarithmic terms. The derivative of log(x) is 1/x, and the derivative of log(2) is 0 since it's a constant.

Putting it all together, the derivative of f(x) is:

f'(x) = 0 + 0 + 1/x + 0

Simplifying further, we get:

f'(x) = 1/x

b) The function l(x) = cot(5ex + 1) can be expressed as a composition of two functions, I and O. In this case, I is the inner function, which is 5ex + 1, and O is the outer function, which is cot(x).

To find the derivative of l(x) using the chain rule, we can use the following notation:

l'(x) = (dO/dI) * (dI/dx)

Using the chain rule, the first part (dO/dI) is the derivative of cot(x) with respect to its inner function, which is -csc^2(x).

The second part (dI/dx) is the derivative of 5ex + 1, which is 5e^(5ex + 1).

Multiplying these two parts together, we get:

l'(x) = -csc^2(5ex + 1) * 5e^(5ex + 1)

c) The function g(x) = tan^(-1)(5x) + cot^(-1)(5x) can also be expressed as a composition of two functions, I and O. Let's break it down:

For the first part, tan^(-1)(5x), I is the inner function, which is 5x, and O is the outer function, which is tan^(-1)(x). Applying the chain rule, we have:

(dO/dI) = 1/(1 + (5x)^2) = 1/(1 + 25x^2)

For the second part, cot^(-1)(5x), I is the inner function, which is 5x, and O is the outer function, which is cot^(-1)(x). Applying the chain rule again, we have:

(dO/dI) = -1/(1 + (5x)^2) = -1/(1 + 25x^2)

Now, let's find the derivative of g(x) using the chain rule:

g'(x) = (dO/dI) * (dI/dx) + (dO/dI) * (dI/dx)

g'(x) = (1/(1 + 25x^2)) * 5 + (-1/(1 + 25x^2)) * 5

Simplifying further, we get:

g'(x) = 5/(1 + 25x^2) - 5/(1 + 25x^2)

g'(x) = 0

d) For the function k(x) = ln(2x - 1) + e^(5x) - 13x + 4, let's break it down into its composition of functions:

I is the inner function for ln(2x - 1), which is 2x - 1, and O is the outer function, which is ln(x).

For e^(5x), I is the inner function, which is 5x, and O is the outer function, which is e^x.

Now, let's apply the chain rule to find the derivative of k(x):

k'(x) = (dO/dI) * (dI/dx) + (dO/dx)

For the first part, (dO/dI) for ln(x) is 1/x, and (dI/dx) for 2x - 1 is 2.

For the second part, (dO/dx) for e^x is e^x.

Putting it all together, we have:

k'(x) = (1/(2x - 1)) * 2 + e^(5x) - 13

Simplifying further, we get:

k'(x) = 2/(2x - 1) + e^(5x) - 13