Wikipedia
derivatives of functions : In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much a quantity is changing at a given point. For example, the derivative of the position (or distance) of a vehicle with respect to time is the instantaneous velocity (respectively, instantaneous speed) at which the vehicle is traveling. Conversely, the integral of the velocity over time is the vehicle's position. The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value. For a real-valued function of a single real variable, the derivative at a point equals the slope of the tangent line to the graph of the function at that point. In higher dimensions, the derivative of a function at a point is a linear transformation called the linearization. A closely related notion is the differential of a function. The process of finding.... More from Wikipedia
derivatives of functions : In calculus, a branch of mathematics, the derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much a quantity is changing at a given point. For example, the derivative of the position (or distance) of a vehicle with respect.. More from Wikipedia
Derivative of a Function
Derivative of a Function - So far we have discussed the derivative of a function f(x) at a point 'a' which is in the domain of f. Suppose we want to find the derivative of the same function at a different point 'b', then we have to compute..
Derivative of a Function of a Function
Derivative of a Function of a Function - So far, we know how to differentiate functions like sin x and x 3 - 5. But how do we differentiate a function of a function? That is how can we differentiate sin (x 3 - 5)?So far, we know how to dif..
Derivative of a Function of a Function - So far, we know how to differentiate functions like sin x and x 3 - 5. But how do we differentiate a function of a function? That is how can we differentiate sin (x 3 - 5)?So far, we know how to dif.. Derivatives of Exponential Functions - I give the basic formulas and do a few examples involving derivatives of exponential functions. For more free math videos, visit JustMathTutoring.com
Inverse Trigonometric Functions - Derivatives - I give the formulas for the derivatives of the six inverse trig functions and do 3 derivative examples. For more free math videos, visit JustMathTutoring.com
Question : I need the derivatives of functions like
Sin(x) = Cos (X)
Cos(x) = -sin(x)
-sin(x) = -cos (x)
-cos(x) = sin(x)
But I need to know the derivatives for tan, cot, csc, and sec.
Answer : cscx = -cscx * cotx secx = secx * tanx cotx = -csc^2(x) tanx = sec^2(x).. More from Yahoo Answers
Answer : cscx = -cscx * cotx secx = secx * tanx cotx = -csc^2(x) tanx = sec^2(x).. More from Yahoo Answers
Question : so, on my math homework, there's an equation x^2+xy+y^2=3. we're currently doing derivatives of implicit functions, but i just don't understand how the process of figuring out what the derivative is as it's stated in the book. can someone give me a little help and lay out the steps to solving this with the given equation?
thanks!
Answer : We differentiate both sides separately. I put a d/dx notation so I won't get lost, although you can skip this part and go straight to differentiating. So we have: d/dx(x + xy + y ) = d/dx(3) We differentiate normally. However, since we're differentiating with respect to x, we must attach dy/dx when differentiating y variables only. Take note that we must use the product rule in differentiating xy. Solution: d/dx(x + xy + y ) = d/dx(3) 2x + (1)y + x(1)(dy/dx) + 2y(dy/dx) = 0 2x + y + x(dy/dx) + 2y(dy/dx) = 0 Transpose 2x and y. x(dy/dx) + 2y(dy/dx) = -2x - y (Factor dy/dx out) dy/dx(x + 2y) = -2x - y (Divide both sides by x + 2y) dy/dx = -2x - y / (x + 2y) (Answer) Notice we have now isolated dy/dx on one side of the equation; hence, we have successfully differentiated implicitly. Hope this helps!.. More from Yahoo Answers
Answer : We differentiate both sides separately. I put a d/dx notation so I won't get lost, although you can skip this part and go straight to differentiating. So we have: d/dx(x + xy + y ) = d/dx(3) We differentiate normally. However, since we're differentiating with respect to x, we must attach dy/dx when differentiating y variables only. Take note that we must use the product rule in differentiating xy. Solution: d/dx(x + xy + y ) = d/dx(3) 2x + (1)y + x(1)(dy/dx) + 2y(dy/dx) = 0 2x + y + x(dy/dx) + 2y(dy/dx) = 0 Transpose 2x and y. x(dy/dx) + 2y(dy/dx) = -2x - y (Factor dy/dx out) dy/dx(x + 2y) = -2x - y (Divide both sides by x + 2y) dy/dx = -2x - y / (x + 2y) (Answer) Notice we have now isolated dy/dx on one side of the equation; hence, we have successfully differentiated implicitly. Hope this helps!.. More from Yahoo Answers
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