Quotient rule

[ "article:topic", "quotient rule", "power rule", "product rule", "change-of-base formula", "authorname:openstax", "product rule of logarithms", "quotient rule of logarithms", "power rule for logarithms", "license:ccby", "showtoc:no", "transcluded:yes", "source[1]-math-15081", "licenseversion:40", "source@https://openstax.org/details/books/precalculus" ] \( \newcommand\) • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Learning Objectives • Use the product rule for logarithms. • Use the quotient rule for logarithms. • Use the power rule for logarithms. • Expand logarithmic expressions. • Condense logarithmic expressions. • Use the change-of-base formula for logarithms. In chemistry, • Battery acid: \(0.8\) • Stomach acid: \(2.7\) • Orange juice: \(3.3\) • Pure water: \(7\) at \(25^\circ C\) • Human blood: \(7.35\) • Fresh coconut: \(7.8\) • Sodium hydroxide (lye): \(14\) To determine whether a solution is acidic or alkaline, we find its pH, which is a measure of the number of active positive hydrogen ions in the solution. The pH is defined by the following formula, where \([\ce\): The pH of hydrochloric acid is tested with litmus paper. (credit: David Berardan). Using the Product Rule for Logarithms Recall that the logarithmic and exponential functions “undo” each other. This means that logarithms have similar properties to exponents. Some important properties of logarithms are given here. First, the following properties a...

Because we originally had f(x)/g(x) which is =f(x)*1/g(x)=f(x)*[g(x)]^-1, that is why 1/g(x)=g(x)^-1, and actually this is true for any number x except zero. So 1/x=x^-1 and similarly we can substitute in any value for x except zero. 1/2=2^-1 1/3=3^-1 1/777=777^-1 If you don't believe me, try it out using a calculator ;-) Hope this helps :-) Be careful - the only multiplication going on in that problem is the "ax" part. This is not a product rule problem. This is a chain rule, within a chain rule problem. The rule remains the same, you just have to do it twice: differentiate the outermost function, keep the inside the same, then multiply by the derivative of the inside. = sec^2[ ln (ax + b) ] * d/dx[ ln (ax + b] = sec^2[ ln (ax + b) ] * (ax + b)^-1 * d/dx (ax + b) = sec^2[ ln (ax + b) ] * (ax + b)^-1 * a The quotient rule could be seen as an application of the product and chain rules. If Q(x) = f(x)/g(x), then Q(x) = f(x) * 1/(g(x)) . You can use the product rule to differentiate Q(x), and the 1/(g(x)) can be differentiated using chain rule with u = g(x), and 1/(g(x)) = 1/u. This is what Sal does in the video. Why this " ´(f(x)/g(x)) = f ´(x) * 1/g(x) + f(x) * 1/g ´(x) " do not works? I understand that Sal is Using the product rule and then the chain rule, but I just do not understand why using just the product rule do not works. I can not explain my doubt better I hope someone could understand my concern and explain me why using the product rule and the chain rule to solv...

Quotient Rule Quotient rule in calculus is a method to find the derivative or differentiation of a function given in the form of a ratio or division of two differentiable functions. That means, we can apply the quotientrule when we have to find the derivative of a function of the form: f(x)/g(x), such that both f(x) and g(x) are differentiable, and g(x) ≠ 0. The quotient rule follows the product rule and the concept of limits of derivation in differentiation directly. Let us understand the formula for quotient rule, its proof using solved examples in detail in the following sections. 1. 2. 3. 4. 5. What is the Quotient Rule? Quotient rule in calculus is a method used to find the derivativeof any function given in the form of a quotient obtained from the result of the division of two differentiable functions. The quotient rule in words states that the derivative of a quotient is equal to the ratio of the result obtained on the subtraction of the numerator times the derivative of the f'(x) = [u(x)/v(x)]' = [v(x) × u'(x) - u(x) × v'(x)]/[v(x)] 2 Quotient Rule Formula We can calculate the derivative or evaluate the f'(x) = [u(x)/v(x)]' = [v(x) × u'(x) - u(x) × v'(x)]/[v(x)] 2 where, • f(x) = The function of the form u(x)/v(x) for which the derivative is to be calculated. • u(x) = A differentiable function that makes numerator of the function f(x). • u'(x) = Derivative of function u(x). • v(x) = A differentiable function that makes denominator of the given function f(x). • v'(x...

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Formula for the derivative of a ratio of functions In quotient rule is a method of finding the h ( x ) = f ( x ) g ( x ) , then using the quotient rule: d d x ( e x x 2 ) = ( d d x e x ) ( x 2 ) − ( e x ) ( d d x x 2 ) ( x 2 ) 2 = ( e x ) ( x 2 ) − ( e x ) ( 2 x ) x 4 = x 2 e x − 2 x e x x 4 = x e x − 2 e x x 3 = e x ( x − 2 ) x 3 . as follows: d d x tan ⁡ x = d d x ( sin ⁡ x cos ⁡ x ) = ( d d x sin ⁡ x ) ( cos ⁡ x ) − ( sin ⁡ x ) ( d d x cos ⁡ x ) cos 2 ⁡ x = ( cos ⁡ x ) ( cos ⁡ x ) − ( sin ⁡ x ) ( − sin ⁡ x ) cos 2 ⁡ x = cos 2 ⁡ x + sin 2 ⁡ x cos 2 ⁡ x = 1 cos 2 ⁡ x = sec 2 ⁡ x . Reciprocal rule [ ] h ′ ( x ) = d d x [ 1 g ( x ) ] = 0 ⋅ g ( x ) − 1 ⋅ g ′ ( x ) g ( x ) 2 = − g ′ ( x ) g ( x ) 2 . added and subtracted to allow splitting and factoring in subsequent steps without affecting the value: h ′ ( x ) = lim k → 0 h ( x + k ) − h ( x ) k = lim k → 0 f ( x + k ) g ( x + k ) − f ( x ) g ( x ) k = lim k → 0 f ( x + k ) g ( x ) − f ( x ) g ( x + k ) k ⋅ g ( x ) g ( x + k ) = lim k → 0 f ( x + k ) g ( x ) − f ( x ) g ( x + k ) k ⋅ lim k → 0 1 g ( x ) g ( x + k ) = lim k → 0 [ f ( x + k ) g ( x ) − f ( x ) g ( x ) + f ( x ) g ( x ) − f ( x ) g ( x + k ) k ] ⋅ 1 g ( x ) 2 = [ lim k → 0 f ( x + k ) g ( x ) − f ( x ) g ( x ) k − lim k → 0 f ( x ) g ( x + k ) − f ( x ) g ( x ) k ] ⋅ 1 g ( x ) 2 = [ lim k → 0 f ( x + k ) − f ( x ) k ⋅ g ( x ) − f ( x ) ⋅ lim k → 0 g ( x + k ) − g ( x ) k ] ⋅ 1 g ( x ) 2 = f ′ ( x ) g ( x ) − f ( x ) g ′ ( x ) g ( x ) 2 . gives: h ′ ( x ) =...