I am interpreting this as:
Find the derivative of:
y=((2x+5)8+4x)3
To find the derivative (y'), the chain rule must be applied. The "outermost" function of this compound function is t3 (t being an arbitrary quantity). The derivative of t3 is 3t2 * dt, where "dt" is the derivative of the quantity "t". Applying this, we arrive at a working definition of y':
y' = 3((2x+5)8+4x)2(derivative of (2x+5)8+4x)
The derivative of (2x+5)8+4x is found using basic derivative definitions and the chain rule again:
8(2x+5)7(2)+4 = 16(2x+5)7+4
So now we can write y' again:
y'= 3((2x+5)8+4x)2(16(2x+5)7+4) = 48((2x+5)8+4x)2((2x+5)7+4)
This can be further simplified, but this is an arduous process. If you need further simplification, feel free to contact me via private message.
14x
y is a sum of constants and so is itself a constant. Its derivative is, therefore, zero.
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
if y=aex+be2x+ce3x then its derivative dy/dx or y' is: y'=aex+2be2x+3ce3x
Rprime= 2q + 2a + 3 Rdoubleprime= 4
x = 10x, so derivative = 10
14x
y is a sum of constants and so is itself a constant. Its derivative is, therefore, zero.
The first derivative is m and the second is 0 so the third is also 0.
Following the correct order of operations: derivative of x^2 + 6/2 = derivative of x^2 +3, which equals 2x
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
if y=aex+be2x+ce3x then its derivative dy/dx or y' is: y'=aex+2be2x+3ce3x
y"+y'=0 is a differential equation and mean the first derivative plus the second derivative =0.Look at e-x the first derivative is -e-xThe second derivative will be e-xThe sum will be 0
Your answer will depend on the parameters of the instructions. If you're looking for the first derivative, simply use the product rule by changing the denominator to a negative exponent and bringing it up (take the negative square root of the quantity x-2 to the top). Then, follow the rules of calculus and algebra. Wow, that's a mess. Let's see... you get "the quantity x cubed plus 6x squared plus 3x plus 1 times the quantity -1(x-2) raised to the negative second plus the quantity x-2 raised to the negative first times the quantity 3x squared plus 12x plus 3." This is because of the Product Rule. Simplifying (by factoring out (x-2) raised to the negative second and combining like terms) gives us "(x-2) raised to the negative second times the quantity 2x cubed minus 24x minus 7." This can also be written as "2x cubed minus 24x minus 7 all over the quantity x-2 squared." f'(x)= 2x^3-24x-7 (x-2)^2
Rprime= 2q + 2a + 3 Rdoubleprime= 4
x = 3.5
irdk