Use implicit differentiation to find (d y)/(d x) at the given point P. y cos^-1(x y)=(-3 √(2))/(

Use implicit differentiation to find (d y)/(d x) at the...

Key Concepts

Implicit Differentiation

Implicit differentiation is a technique used when a function is defined implicitly rather than explicitly. It involves differentiating both sides of the equation with respect to x while treating y as a function of x, which then allows for solving for dy/dx even when y is not isolated.

Chain Rule

The chain rule is essential for differentiating composite functions, where one function is nested within another. In implicit differentiation, it is used to differentiate expressions where y, or a function of y, is composed with another function, ensuring that the derivative of the inner function is appropriately accounted for.

Product Rule

The product rule is applied when differentiating an expression that is the product of two functions. When an equation involves the multiplication of y with another function of x and y, the product rule helps in correctly differentiating each component and combining the resulting derivatives.

Differentiation of Inverse Trigonometric Functions

Differentiating inverse trigonometric functions requires the use of specific derivative formulas. For instance, the derivative of arccos(u) with respect to x is given by -u' divided by the square root of (1 - u^2). This concept is particularly important when the implicit equation includes inverse trigonometric expressions.

Evaluating Derivatives at a Specific Point

After obtaining a general expression for dy/dx using implicit differentiation, it is often necessary to evaluate the derivative at a particular point. This involves substituting the given x and y values into the derivative expression to compute the numerical slope of the tangent at that point.

Transcript

00:01 The equation of the form y times cosine inverse of xy and equal to minus 2 squared of 2 of 4 pi and we are given the point b a half and minus square of 2 and we need to find the y prime at the p here so given the point p here we know the value of the x and the y here and in the channel prime we need to use the implicit derivative to the derivative on both sides here on the left hand side notice that we have two functions therefore we need to apply the product rule u times v derivative equal to u prime v plus u v prime so using this formula we should get now the left hand side will become y prime 10 cosine inverse of xy and now plus y times cosine inverse of the xy derivative and on the right hand side will equal to it's a constant therefore equal to zero now the next step for the y brand we don't is the one we're looking for so i will keep the first term exactly as it is now for the second term the derivative of the cosay inverse by formula which is equal to the xy prime divided by square root of the one minus xy square but this is a minus therefore have to be the minus now equal to zero and the last thing for this one is the again is a derivative of the product so one more step we get the y prime go say inverse of the xy minus y and then times with the on the time we have the x pram will be equal to 1 by product rule so we have a 1 will be y plus x y prime divided by square root of the 1 minus x y square equal to 0 and now we are ready to blake in the value of the x and the y here so x equal to a half y equal to minus a square root of 2 therefore we should get now y prime cosine inverse of x times y will get equal to the minus square of two over two minus y equal to minus square of two so plus square of two and now on the top we have the minus square root of two and plus the x equal to a half so it will be a half of the y prime divided by square root of of 1 minus x times y square so we have it will equal to x times y equal to minus squared up and squared up we get the 2 over 4 so equal to a half so we get here we equal to a half here equal to 0 and now we can get there for this one cosine of the minus square of two out of two will equal to three by after four plus now for this one we got the one minus square one minus a half will be yeah it will equal to a half and the square root to be one number square of two so we should get one number square of two with this square of two we get the two now and now times with the minus square root of 2 plus 1 half y prime equal to 0.

04:01 The next step we'll distribute the 2 inside.

04:04 Therefore we get the y prime and write the constant first, 3x4, y prime, and plus y prime minus 2 square root 2 equal to 0...

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