Differentiate the following function.
f (x) = 6 x 2 sin(x)tan(x)
Where do I begin?
f (x) = 6 x 2 sin(x)tan(x)
begin with
12 [ sin x d/dx (tan x) + tan x d/dx(sin x)]
Begin by factoring out the 12.
f'(x) = 12 d/dx (sin x tan x)
Then use the product rule
f'(x) = 12[tanx d/dx(sinx) + sinx d/dx(tanx)]
= 12 (tanx*cosx + sinx*sec^2 x)
= 12 (sinx + secx tanx)
Check my steps. I don't have my books handy.
To differentiate the function f(x) = 6x^2 sin(x)tan(x), you can use the product rule and the chain rule.
The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is given by the formula (u·v)' = u'v + uv'.
In this case, we can let u(x) = 6x^2 and v(x) = sin(x)tan(x). Now we can find the derivatives of u(x) and v(x) respectively.
To differentiate u(x) = 6x^2, we can use the power rule, which states that if you have a function of the form f(x) = x^n, then its derivative is given by the formula f'(x) = nx^(n-1). Applying this rule, we have u'(x) = 2(6)x^(2-1) = 12x.
To differentiate v(x) = sin(x)tan(x), we need to use both the chain rule and the product rule. Let's break it down further:
Let's define w(x) = sin(x) and z(x) = tan(x).
Now, applying the product rule, we can find the derivative of v(x) as follows:
v'(x) = w'(x)z(x) + w(x)z'(x)
To find w'(x) and z'(x), we need to differentiate each function separately:
For w(x) = sin(x), we can differentiate it using the chain rule. The chain rule states that if you have a function f(g(x)), then its derivative is given by the formula f'(g(x)) · g'(x). Here, f(x) = sin(x) and g(x) = x. So, applying the chain rule, we have w'(x) = cos(x)·1 = cos(x).
For z(x) = tan(x), we can differentiate it using the quotient rule, which states that if you have a function of the form f(x) = g(x)/h(x), then its derivative is given by the formula f'(x) = (g'(x)h(x) - g(x)h'(x))/[h(x)]^2. In this case, g(x) = sin(x) and h(x) = cos(x). Applying the quotient rule, we have z'(x) = (cos(x)·cos(x) - sin(x)·(-sin(x)))/[cos(x)]^2 = (cos^2(x) + sin^2(x))/cos^2(x) = 1/cos^2(x) = sec^2(x).
Now that we have w'(x) = cos(x) and z'(x) = sec^2(x), we can substitute them back into the equation for v'(x):
v'(x) = w'(x)z(x) + w(x)z'(x)
= cos(x)·tan(x) + sin(x)·sec^2(x)
Finally, substituting this into the product rule for f(x), we have:
f'(x) = u'(x)v(x) + u(x)v'(x)
= 12x·(sin(x)tan(x)) + (6x^2)·(cos(x)·tan(x) + sin(x)·sec^2(x))
And this is the derivative of the given function f(x) = 6x^2 sin(x)tan(x).