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What else can I help you with?
Find the derivative of y equals 3cosx?
y=3 cos(x) y' = -3 sin(x)
What is the derivative of sinπx?
The derivative with respect to 'x' of sin(pi x) ispi cos(pi x)
Find y'' if y equals 6x sinx?
That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
What is the derivative of sin x plus 2?
d/dx [sin(x) + 2] = cos(x)
What is the derivative of sin x power 0?
I think you are asking "what is the derivative of [sin(x)]^0=sin^0(x)?" and I shall answer this accordingly. Recall that x^0 = 1 whenever x is not 0. On the other hand, also notice that 0^0 is generally left undefined. Thus, sin^0(x) is the function f(x) such that f(x) is undefined when x = n(pi) and 1 everywhere else. As a result, on every open interval not containing a multiple of pi, i.e. on (n(pi), (n+1)(pi)) the derivative will be zero, since f is just a constant function on these intervals, and whenever x is a multiple of pi, the derivative at x will be undefined. Thus, [d/dx]sin^0(x) is undefined whenever x = n(pi) and 0 everywhere else. In some cases, mathematicians define 0^0 to be 1, and if we were to use this convention, sin^0(x) = 1 for all x, and its derivative would just be 0.
Find the derivative y equals 10 to the sin3x power?
Y=10^sin(x) The derivative is: (log(5)+log(2))*cos(x)*2^sin(x)*5^sin(x) Use the chain rule, product rule, and power rules combined with sin(x) rule.
What is the 87th derivative of sin x?
Every fourth derivative, you get back to "sin x" - in other words, the 84th derivative of "sin x" is also "sin x". From there, you need to take the derivative 3 more times, getting:85th derivative: cos x86th derivative: -sin x87th derivative: -cos x
What is the derivative of root sin2x?
You are supposed to use the chain rule for this. First step: derivative of root of sin2x is (1 / (2 root of sin 2x)) times the derivative of sin 2x. Second step: derivative of sin 2x is cos 2x times the derivative of 2x. Third step: derivative of 2x is 2. Finally, you need to multiply all the parts together.
What is the derivative of 3cosx2?
I'm assuming your question reads "What is the derivative of 3cos(x2)?" You must use the Chain Rule. The derivative of cos(x2) equals -sin(x2) times the derivative of the inside (x2), which is 2x. So... d/dx[3cos(x2)] = -6xsin(x2)
How do you find the derivative of y equals sin8x?
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
What is the derivative of x squared times sin sqaured x?
2xsin2x+2x2sinxcosx
Derivative of cosx?
The derivative of cos(x) is negative sin(x). Also, the derivative of sin(x) is cos(x).
Find the derivative of y equals 3cosx?
y=3 cos(x) y' = -3 sin(x)
When does cos x equal -sin x?
Cos (x) = -Sin(x) 1 = -Sin(x) / Cos (x) 1 = -Tan(x) Tan(x) = -1 x = Tan^-1(-1( x = -45 degrees = - pi /4 , 3pi/4, 5pi/4 ....
What is the derivative of 8cosx?
Since 8 is a constant, you can use the rule for constants.In other words, the derivative is 8 times the derivative of cos x.
Differentiate y equals xsinx plus cosx?
y = x sin(x) + cos(x)Derivative of the first term = x cos(x) + sin(x)Derivative of the second term = -sin(x)y' = Sum of the derivatives = x cos(x) + sin(x) - sin(x)= [ x cos(x) ]
How do you differentiate sin sin x?
To differentiate y=sin(sin(x)) you need to use the chain rule. A common way to remember the chain rule is "derivative of the outside, keep the inside, derivative of the inside". First, you take the derivative of the outside. The derivative of sin is cos. Then, you keep the inside, so you keep sin(x). Then, you multiple by the derivative of the inside. Again, the derivative of sinx is cosx. In the end, you get y'=cos(sin(x))cos(x))
