Find the first derivative for the following functions
1) x^3 . sqrt( 3x^5 + 1 ) . sin (7x)
we can rewrite this as
sqrt(x^6)*sqrt(3x^5 + 1)*sin(7x)
sqrt(3x^11 + x^6)*sin(7x)
(3x^11 + x^6)^(1/2) * sin(7x)
then recall that derivative of two functions of x that are multiplied is equal to,
d/dx ( f(x)*g(x) ) = f'(x)g(x) + f(x)g'(x)
therefore,
(1/2)*(33x^10 + 6x^5)*(3x^11 + x^6)^(-1/2)*(sin 7x) + (3x^11 + x^6)^(1/2)*(7 cos 7x)
hope this helps~ :)
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To find the first derivative of the given function, we can apply the product rule and chain rule as necessary. Here's how we can do it step by step:
Step 1: Apply the product rule.
Let's denote the given function as f(x) = x^3 * sqrt(3x^5 + 1) * sin(7x).
To apply the product rule, we can break it into three separate parts: g(x) = x^3, h(x) = sqrt(3x^5 + 1), and k(x) = sin(7x).
Step 2: Find the derivative of each individual part.
The derivative of g(x) = x^3 is g'(x) = 3x^2 (using the power rule).
To find the derivative of h(x), we need to apply the chain rule. Let u(x) = 3x^5 + 1. We have h(x) = sqrt(u(x)). Applying the chain rule, we get h'(x) = (1/2) * (1/sqrt(u(x))) * u'(x).
To find u'(x), we differentiate u(x) using the power rule: u'(x) = 15x^4.
Therefore, h'(x) = (1/2) * (1/sqrt(u(x))) * u'(x) = (1/2) * (1/sqrt(3x^5 + 1)) * 15x^4.
Lastly, the derivative of k(x) = sin(7x) is k'(x) = 7 cos(7x) (using the chain rule).
Step 3: Apply the product rule formula for f(x).
Using the product rule, the first derivative of f(x) can be found as follows:
f'(x) = g'(x) * h(x) * k(x) + g(x) * h'(x) * k(x) + g(x) * h(x) * k'(x).
Applying the formula:
f'(x) = (3x^2) * (sqrt(3x^5 + 1)) * (sin(7x)) + (x^3) * ((1/2) * (1/sqrt(3x^5 + 1)) * 15x^4) * (sin(7x)) + (x^3) * (sqrt(3x^5 + 1)) * (7 cos(7x)).
Simplifying this expression will yield the first derivative of the given function, f(x) = x^3 * sqrt(3x^5 + 1) * sin(7x).