Sin inverse formula

3 Derivatives • Introduction • 3.1 Defining the Derivative • 3.2 The Derivative as a Function • 3.3 Differentiation Rules • 3.4 Derivatives as Rates of Change • 3.5 Derivatives of Trigonometric Functions • 3.6 The Chain Rule • 3.7 Derivatives of Inverse Functions • 3.8 Implicit Differentiation • 3.9 Derivatives of Exponential and Logarithmic Functions • 4 Applications of Derivatives • Introduction • 4.1 Related Rates • 4.2 Linear Approximations and Differentials • 4.3 Maxima and Minima • 4.4 The Mean Value Theorem • 4.5 Derivatives and the Shape of a Graph • 4.6 Limits at Infinity and Asymptotes • 4.7 Applied Optimization Problems • 4.8 L’Hôpital’s Rule • 4.9 Newton’s Method • 4.10 Antiderivatives • 5 Integration • Introduction • 5.1 Approximating Areas • 5.2 The Definite Integral • 5.3 The Fundamental Theorem of Calculus • 5.4 Integration Formulas and the Net Change Theorem • 5.5 Substitution • 5.6 Integrals Involving Exponential and Logarithmic Functions • 5.7 Integrals Resulting in Inverse Trigonometric Functions • 6 Applications of Integration • Introduction • 6.1 Areas between Curves • 6.2 Determining Volumes by Slicing • 6.3 Volumes of Revolution: Cylindrical Shells • 6.4 Arc Length of a Curve and Surface Area • 6.5 Physical Applications • 6.6 Moments and Centers of Mass • 6.7 Integrals, Exponential Functions, and Logarithms • 6.8 Exponential Growth and Decay • 6.9 Calculus of the Hyperbolic Functions • Learning Objectives • 1.4.1 Determine the conditions for when a ...

Good question! Sal skipped over the reason, and it's far from obvious. The answer has to do with the way we define the inverse sine function. As you know, the sine function is cyclical, so we wouldn't get a valid function if we created its inverse in the usual way of simply reflecting it across the y = x diagonal because we wouldn't have a unique output for valid inputs. For example, the inverse sine of 0 could be 0, or π, or 2π, or any other integer multiplied by π. To solve this problem, we restrict the range of the inverse sine function, from -π/2 to π/2. Within this range, the slope of the tangent is always positive (except at the endpoints, where it is undefined). Therefore, the derivative of the inverse sine function can't be negative. We can safely discard the negative square root in this derivation because it would give us a negative derivative, which is impossible because of the way we've restricted the domain of the inverse sine function. I am a bit rusty with taking derivatives of inverse functions but is there a reason why when taking the derivative of y=(sin x)^-1 that we could not use the Power or Quotient Rule to solve the function? When I use the power/quotient rule I get d(y)/dx = d((sin x)^-1)/dx = -cos x/(sin x)^2 and not 1/sqrt(1-x^2) that Sal mentions. Where am I going wrong? The inverse functions, though written as sin⁻¹, etc. ARE NOT the reciprocals of those functions. They are NOT being raised to the -1 power. Thus, what you were doing was finding t...

"Sine" and "Cosine" redirect here. For other uses, see Sine and cosine General information General definition sin ⁡ ( α ) = opposite hypotenuse cos ⁡ ( α ) = adjacent hypotenuse Fields of application • • • • • Reference • • • • Laws and theorems • • • • • • • • • v • t • e In sine and cosine are θ . More generally, the definitions of sine and cosine can be extended to any The sine and cosine functions are commonly used to model jyā and koṭi-jyā functions used in Notation [ ] Main article: Sine and cosine are written using sin and cos. Often, if the argument is simple enough, the function value will be written without parentheses, as sin θ rather than as sin( θ). Each of sine and cosine is a function of an angle, which is usually expressed in terms of Definitions [ ] Right-angled triangle definitions [ ] α, the sine function gives the ratio of the length of the opposite side to the length of the hypotenuse. To define the sine and cosine of an acute angle α, start with a α; in the accompanying figure, angle α in triangle ABC is the angle of interest. The three sides of the triangle are named as follows: • The opposite side is the side opposite to the angle of interest, in this case side a. • The hypotenuse is the side opposite the right angle, in this case side h. The hypotenuse is always the longest side of a right-angled triangle. • The adjacent side is the remaining side, in this case side b. It forms a side of (and is adjacent to) both the angle of interest (angle A) a...

3 Derivatives • Introduction • 3.1 Defining the Derivative • 3.2 The Derivative as a Function • 3.3 Differentiation Rules • 3.4 Derivatives as Rates of Change • 3.5 Derivatives of Trigonometric Functions • 3.6 The Chain Rule • 3.7 Derivatives of Inverse Functions • 3.8 Implicit Differentiation • 3.9 Derivatives of Exponential and Logarithmic Functions • 4 Applications of Derivatives • Introduction • 4.1 Related Rates • 4.2 Linear Approximations and Differentials • 4.3 Maxima and Minima • 4.4 The Mean Value Theorem • 4.5 Derivatives and the Shape of a Graph • 4.6 Limits at Infinity and Asymptotes • 4.7 Applied Optimization Problems • 4.8 L’Hôpital’s Rule • 4.9 Newton’s Method • 4.10 Antiderivatives • 5 Integration • Introduction • 5.1 Approximating Areas • 5.2 The Definite Integral • 5.3 The Fundamental Theorem of Calculus • 5.4 Integration Formulas and the Net Change Theorem • 5.5 Substitution • 5.6 Integrals Involving Exponential and Logarithmic Functions • 5.7 Integrals Resulting in Inverse Trigonometric Functions • 6 Applications of Integration • Introduction • 6.1 Areas between Curves • 6.2 Determining Volumes by Slicing • 6.3 Volumes of Revolution: Cylindrical Shells • 6.4 Arc Length of a Curve and Surface Area • 6.5 Physical Applications • 6.6 Moments and Centers of Mass • 6.7 Integrals, Exponential Functions, and Logarithms • 6.8 Exponential Growth and Decay • 6.9 Calculus of the Hyperbolic Functions • Learning Objectives • 1.4.1 Determine the conditions for when a ...

"Sine" and "Cosine" redirect here. For other uses, see Sine and cosine General information General definition sin ⁡ ( α ) = opposite hypotenuse cos ⁡ ( α ) = adjacent hypotenuse Fields of application • • • • • Reference • • • • Laws and theorems • • • • • • • • • v • t • e In sine and cosine are θ . More generally, the definitions of sine and cosine can be extended to any The sine and cosine functions are commonly used to model jyā and koṭi-jyā functions used in Notation [ ] Main article: Sine and cosine are written using sin and cos. Often, if the argument is simple enough, the function value will be written without parentheses, as sin θ rather than as sin( θ). Each of sine and cosine is a function of an angle, which is usually expressed in terms of Definitions [ ] Right-angled triangle definitions [ ] α, the sine function gives the ratio of the length of the opposite side to the length of the hypotenuse. To define the sine and cosine of an acute angle α, start with a α; in the accompanying figure, angle α in triangle ABC is the angle of interest. The three sides of the triangle are named as follows: • The opposite side is the side opposite to the angle of interest, in this case side a. • The hypotenuse is the side opposite the right angle, in this case side h. The hypotenuse is always the longest side of a right-angled triangle. • The adjacent side is the remaining side, in this case side b. It forms a side of (and is adjacent to) both the angle of interest (angle A) a...

Good question! Sal skipped over the reason, and it's far from obvious. The answer has to do with the way we define the inverse sine function. As you know, the sine function is cyclical, so we wouldn't get a valid function if we created its inverse in the usual way of simply reflecting it across the y = x diagonal because we wouldn't have a unique output for valid inputs. For example, the inverse sine of 0 could be 0, or π, or 2π, or any other integer multiplied by π. To solve this problem, we restrict the range of the inverse sine function, from -π/2 to π/2. Within this range, the slope of the tangent is always positive (except at the endpoints, where it is undefined). Therefore, the derivative of the inverse sine function can't be negative. We can safely discard the negative square root in this derivation because it would give us a negative derivative, which is impossible because of the way we've restricted the domain of the inverse sine function. I am a bit rusty with taking derivatives of inverse functions but is there a reason why when taking the derivative of y=(sin x)^-1 that we could not use the Power or Quotient Rule to solve the function? When I use the power/quotient rule I get d(y)/dx = d((sin x)^-1)/dx = -cos x/(sin x)^2 and not 1/sqrt(1-x^2) that Sal mentions. Where am I going wrong? The inverse functions, though written as sin⁻¹, etc. ARE NOT the reciprocals of those functions. They are NOT being raised to the -1 power. Thus, what you were doing was finding t...