can some one please help me with the following questions as i don't understand them please
(a) write down the derivative of each of the functions
f(x)=E7x and g(x)=cos(2x)
using the product rule differentiate the function
k(x)=E7xcos(2x)
(b)write down the derivative of each of the functions
f(t)=t5-1 and g(t)=1n(4t) (t>0)
using the quotient rule differentiate the function
k(t)=t5-1 (t>1)
- -
1n(4t) 4
(c) write down the derivative of each of the functions
f(x)=x3/2 (x>0) and g(u)=sin(1u)
-
3
useing the composite rule differentiate the function
k(x)=sin(1x3/2) (x>0)
-
3
Sure, I'd be happy to help you with these questions! Let's break them down one by one.
(a) For the first part, we need to find the derivative of two functions: f(x) = e^(7x) and g(x) = cos(2x).
To find the derivative of f(x), we can use the chain rule. The derivative of e^(ax) is a * e^(ax). So, the derivative of f(x) = e^(7x) is f'(x) = 7 * e^(7x).
Next, let's find the derivative of g(x) = cos(2x). The derivative of cos(x) is -sin(x), and we need to apply the chain rule here as well. Since the inside function is 2x, the derivative of g(x) = cos(2x) is g'(x) = -sin(2x) * 2.
Now, using the product rule, we can differentiate k(x) = f(x) * g(x). The product rule states that the derivative of the product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.
So, applying the product rule, the derivative of k(x) = e^(7x) * cos(2x) is k'(x) = (7 * e^(7x)) * cos(2x) + e^(7x) * (-sin(2x) * 2).
(b) Moving on to the next part, we have f(t) = t^5 - 1 and g(t) = ln(4t). We need to find their derivatives using the quotient rule to differentiate k(t) = (t^5 - 1) / ln(4t).
The quotient rule states that the derivative of the quotient of two functions is (the derivative of the numerator times the denominator) minus (the numerator times the derivative of the denominator), all divided by (the denominator squared).
To find the derivative of f(t) = t^5 - 1, we can simply apply the power rule. The derivative of t^n is n * t^(n-1). So, the derivative of f(t) = t^5 - 1 is f'(t) = 5 * t^4.
Next, we can find the derivative of g(t) = ln(4t) using the chain rule. The derivative of ln(x) is 1/x, so the derivative of g(t) = ln(4t) is g'(t) = 1/(4t) * 4 = 1/t.
Now, using the quotient rule, the derivative of k(t) = (t^5 - 1) / ln(4t) is k'(t) = (5 * t^4 * ln(4t) - (t^5 - 1) * 1/t) / (ln(4t))^2.
(c) In the last part, we have f(x) = x^(3/2) and g(u) = sin(u). We need to differentiate k(x) = sin(x^(3/2)) / 3.
To differentiate f(x) = x^(3/2), we can use the power rule. The power rule states that the derivative of x^n is n * x^(n-1). So, the derivative of f(x) = x^(3/2) is f'(x) = (3/2) * x^(3/2 - 1) = (3/2) * x^(1/2).
Next, we can find the derivative of g(u) = sin(u), which is simply the derivative of sin(x). The derivative of sin(x) is cos(x). So, the derivative of g(u) = sin(u) is g'(u) = cos(u).
Using the composite rule, the derivative of k(x) = sin(x^(3/2)) / 3 is k'(x) = (cos(x^(3/2)) * (3/2) * x^(1/2)) / 3.
I hope this helps! Let me know if you have any further questions.