Hi! I am having trouble figuring out the derivative of ln[sin^2(x)+3]...........
If it weren't for the "3" I know I could just "move" the exponent of 2 from sin^2 to in front of the ln,....but because of the 3 I can't do this....So what should I do?
Thank you very much for your help! Have a nice day :)
Let u = sin^2x + 3
f(x) = ln[sin^2(x)+3]
= ln u(x)
df/dx = df/du*du/dx
= (1/u)*2sinx*cosx
= sin(2x)/[sin^2x+3]
thank you! but how did you go from......(1/u)*2sinx*cosx....to......sin(2x)/[sin^2x+3].............? i was just confused by this..thanks
To find the derivative of ln[sin^2(x) + 3], you can use the chain rule.
Step 1: Let's simplify the expression first. Since sin^2(x) + 3 does not factor easily, we cannot directly move the exponent of 2 from sin^2(x) in front of ln. So, we'll keep it as it is for now.
Step 2: Now, let's consider the function inside the ln: f(x) = sin^2(x) + 3.
Step 3: We need to find the derivative of f(x) with respect to x. The derivative of sin^2(x) is 2sin(x)cos(x) according to the chain rule. The derivative of the constant 3 is zero.
Step 4: The derivative of f(x) is f'(x) = 2sin(x)cos(x) + 0 = 2sin(x)cos(x).
Step 5: Now, using the chain rule, the derivative of ln[f(x)] is given by d(ln[f(x)])/dx = (f'(x))/f(x).
So, the derivative of ln[sin^2(x) + 3] is (2sin(x)cos(x))/(sin^2(x) + 3).
Remember, the chain rule is applied when you have a composition of functions, where in this case, ln is the outer function and sin^2(x) + 3 is the inner function.