Sunday, August 27, 2023
Product Rule & Quotient Rule & Chain Rule (Derivative)
You can find the product rule, quotient rule and chain rule in the derivative below. We wish you good lessons.
Product Rule:
$\frac{d}{dx}(u(x).v(x))=u(x)\frac{dv}{dx}+v(x)\frac{du}{dx}$
Quotient Rule:
$\frac{d}{dx}(\frac{u(x)}{v(x)})=\frac{v(x)\frac{du}{dx}-u(x)\frac{dv}{dx}}{[v(x)]^2}$
Chain Rule:
$\frac{d}{dx}(f(g(x)))=f'(g(x))×g'(x)$
Tuesday, August 22, 2023
Laplace Transforms | Function and Transform (Examples)
You can find Laplace transforms below. We wish everyone good work.
You can find sample solutions below.
Sunday, August 20, 2023
General Rules of Differantiation
In the following, u, v, w are functions of x; a, b, c, n are constants [restricted if indicated]; e=2.71828... is the natural base of logarithms; In u is the natural loraithm of u [i.e. the loraithm to the base e] where it is assumed thad u>0 and all angles are in radians.
Sunday, November 27, 2022
$sec^{-1} x^2$ Derivative | What is derivative of sec x^2?
Hello everyone. You can reach derivative of $sec^{-1} x^2$ in this lesson. We wish you good work.
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sec -1 derivative |
What is derivative of $sec^{-1} x^2$?
$Sec^{-1} u$ derivative formulas:
$\frac{d}{dx}sec^{-1} u=\frac{1}{|u|.\sqrt{u^2-1}}.\frac{du}{dx}$
Now let's answer our question.
Differentiate $y=sec^{-1} x^2$
Solution For $x^2$ > 1 > 0,
>>> $\frac{dy}{dx}=\frac{1}{|x|\sqrt{(x^2)^2-1}}.\frac{d}{dx}x^2$
>>> $=\frac{2x}{x^2\sqrt{x^4-1}}=\frac{2}{x.\sqrt{x^4-1}}$
Answer = $\frac{2}{x.\sqrt{x^4-1}}$