How is 6#sin(#x-1)/cos^3(#x-1) the derivative of f(x)= 3sec^2(#x-1)?
#= Pi (I coulnd't find the symbol...sorry.)
recall that d(secu)/dx
= tanu secu du/dx
so if f(x) = 3sec^2(pix-1)
= 3(sec(pix-1))^2 then using the chain rule
f'(x) = 6(sec(pix-1)*tan(pix-1)sec(pix-1)*pi
= 6pi(sin(pix-1)/cos(pix-1)*sec^2(pix-1)
= 6pi(sin(pix-1))/cos(pix-1)*1/cos^2(pix-1)
= 6pi(sinpix-1)/cos^3(pix-1)
To find the derivative of f(x) = 3sec^2(x-1), you can use the quotient rule.
The quotient rule states that if you have a function of the form u/v, where both u and v are differentiable functions, the derivative can be found by the following formula:
(f/g)' = (g * f' - f * g') / g^2
In this case, u = 6*sin(x-1) and v = cos^3(x-1).
First, we will find the derivative of u with respect to x. Since u is a function of sin(x-1), we can use the chain rule. The chain rule states that if you have a composite function, f(g(x)), the derivative can be found by multiplying the derivative of the outer function with the derivative of the inner function.
Let's start with finding du/dx:
du/dx = 6 * cos(x-1) * (x-1)'
= 6 * cos(x-1)
Next, we will find the derivative of v with respect to x. Since v is a function of cos^3(x-1), we can again use the chain rule.
dv/dx = 3 * cos^2(x-1) * (x-1)'
= 3 * cos^2(x-1)
Now we can apply the quotient rule to find the derivative of f(x):
f'(x) = (v * u' - u * v') / v^2
= (cos^3(x-1) * 6 * cos(x-1) - 6 * sin(x-1) * 3 * cos^2(x-1)) / (cos^3(x-1))^2
Simplifying the expression:
f'(x) = 6 * cos^4(x-1) - 18 * sin(x-1) * cos^2(x-1) / cos^6(x-1)
Since the symbol "#" represents π, we can substitute "#" with π in the expression to get the final answer:
f'(x) = 6 * cos^4(π-1) - 18 * sin(π-1) * cos^2(π-1) / cos^6(π-1)
Therefore, the derivative of f(x) = 3sec^2(π-1) with respect to x is 6 * cos^4(π-1) - 18 * sin(π-1) * cos^2(π-1) / cos^6(π-1).
To find the derivative of the function f(x) = 3sec^2((#x-1)), where # denotes the symbol Pi (π), we can use the chain rule.
The chain rule states that if we have a function of the form g(h(x)), then the derivative of g(h(x)) with respect to x is given by g'(h(x)) * h'(x).
In this case, g(x) = 3x^2 and h(x) = sec((#x-1)).
Let's find the derivatives of g(x) and h(x) separately:
1. Derivative of g(x):
g'(x) = d/dx(3x^2)
= 6x
2. Derivative of h(x):
h'(x) = d/dx(sec((#x-1)))
To find the derivative of sec((#x-1)), we need to apply the chain rule again. The derivative of sec(x) is sec(x)tan(x), so we can express sec((#x-1)) as sec(x)tan(x), with x = (#x-1):
h'(x) = sec(x)tan(x)
Now, substitute x = (#x-1) back into the equation:
h'(x) = sec((#x-1))tan((#x-1))
Therefore, the derivative of f(x) = 3sec^2((#x-1)) is given by:
f'(x) = g'(h(x)) * h'(x)
= 6(#x-1) * sec((#x-1))tan((#x-1))
= 6(#x-1)sec((#x-1))tan((#x-1))
You mentioned a term with sin and cos in your question, but there's no sin or cos functions involved in the derivative of f(x) = 3sec^2((#x-1)). So, there seems to be some confusion or mistyping in your expression 6#sin(#x-1)/cos^3(#x-1).