vela said:It would help if you clarified exactly what the function is. Your notation is a bit ambiguous. Besides the question about the log that Raskolnikov raised, is the first term the arcsin or is it the reciprocal of sin? That notation usually means arcsin 3x, but you've interpreted it as 1/sin 3x.
Raskolnikov said:I love WolframAlpha. There's an option to show the solution for the derivation. Just click the option "show steps" located on the upper-right corner of the Derivative box where it displays the answer.
Char. Limit said:Sometimes W|A integrates oddly though... I once saw it do four u-subs to the same variable chain, when a simple parts integral would suffice.
vela said:The first term means arcsin 3x, not 1/(sin 3x). So try differentiating that.
Differentiating the second term is a bit simpler if you use a properties of the log to rewrite it as
[tex]\log \sqrt{x \sec x} = \log (x \sec x)^{1/2} = \frac{1}{2} \log (x \sec x) = \frac{1}{2} (\log x + \log \sec x)[/tex]
Char. Limit said:Close. The derivative of sec(x) is sec(x)tan(x), not just tan(x), in the last part.
3songs said:OK, so the last part of my equation is
tanx/2secxtanx=1/2secx
?
Thank you!
Char. Limit said:Not quite. Originally, you wrote 1/sec(x) * tan(x). This was close, but it needs to be (1/sec(x))(sec(x)tan(x)).
Consider this little fact:
[tex]\left(log\left(f\left(x\right)\right)\right)'=\frac{f'(x)}{f(x)}[/tex]
So, since f(x)=sec(x), f'(x)=sec(x)tan(x).
3songs said:How silly I am.
After I make these changes, I have:
3/(sqrt(1-(3x)2) +1/(2x) +tanx/2
Is this the final answer?
A derivative is a mathematical concept that represents the rate of change of a function with respect to its input variable. It is essentially the slope of a curve at a specific point.
Derivatives are important in many areas of science and engineering, as they help us understand the behavior of functions and make predictions about their future values. They are especially useful in calculus, where they are used to solve problems involving rates of change and optimization.
The process of finding a derivative involves using mathematical rules and formulas to calculate the slope of a function at a given point. This can be done using the power rule, product rule, quotient rule, or chain rule, depending on the complexity of the function.
Derivatives have a wide range of real-life applications, including predicting the motion of objects in physics, determining the optimal production level in economics, and analyzing the growth of populations in biology. They are also used in engineering, finance, and many other fields.
Although derivatives are incredibly useful, they also have some limitations and challenges. These can include complex functions with multiple variables, non-differentiable points, and the need for precise mathematical calculations. It is important to be aware of these challenges and to use appropriate methods and tools when working with derivatives.