x = yy differentiate both sides with respect to x dx = (y * yy-1) dy dy/dx = y * yy-1 dy/dx = yy = x hence differentiate of y wrt x is x only
y=sinx y=cosxsinx=cosx=>sinx/cosx=1=>tanx=1=>x=45oie.. y=sin45=cos45y=1/(square root of 2)
yes y=sinx is x=arcsiny
The differential of the product xy with respect to x is y + x dy/dx. The differential of logy with respect to x is (1/y) dy/dx. The role of c in this question is not made clear.
y=ax y'=ln(a)*ax
y = (sinx)^(e^x) ln(y) = ln((sinx)^(e^x)) ln(y) = (e^x)ln(sinx) (1/y)dy = (e^x)(1/sinx)(cosx)+ln(sinx)(e^x)dx (1/y)dy = (e^x)(cotx)+ln(sinx)(e^x)dx dy = ((sinx)^(e^x))((cotx)(e^x)+ln(sinx)(e^x))dx dy = ((e^x)(sinx)^(e^x))(cotx+ln(sinx))dx
x = yy differentiate both sides with respect to x dx = (y * yy-1) dy dy/dx = y * yy-1 dy/dx = yy = x hence differentiate of y wrt x is x only
f(x) = 1/x except where x = 0.
y=sinx y=cosxsinx=cosx=>sinx/cosx=1=>tanx=1=>x=45oie.. y=sin45=cos45y=1/(square root of 2)
#include double x, y;y = sin (x);
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
yes y=sinx is x=arcsiny
The differential of the product xy with respect to x is y + x dy/dx. The differential of logy with respect to x is (1/y) dy/dx. The role of c in this question is not made clear.
given the identity sin(x+y)=sinx cosy + siny cosxsin2x = 2 sinx cosx andsin(2(x)+x) = sin 2x cos x + sinx cos 2xusing the last two identities givessin3x= 2 sinx cosx cosx + sinx cos2xfactoring the sinx we havesin3x = sinx(2cosx cosx+cos2x)which satisfies the requirement.However, we can simplify further since cos 2x = cosx cosx - sinx sinx (a well known identity)sin3x = sinx (2cosx cosx +cosx cosx - sinx sinx)so sin3x= sinx(3cosx cosx - sinx sinx)or sin 3x = 3.cos²x.sinx - sin³x* * * * *Good, but not good enough. The answer was required in terms of sin, not a mixture of sinx and cosx. Easily recitified, though, since cos²x = 1 - sin²xTherefore sin3x = 3*(1-sin²x)*sinx - sin³x= 3sinx - 3sin³x - sin³x= 3sinx - 4sin³x
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))
Differentiating with respect to one variable means thinking of all the other variables as constants. The answer is thus y=0.
so say y = 2^x dy/dx = ln2.2^x (. = multiplication symbol, ^ = to-the-power-of symbol) The general formula is (where 'a' is a constant, x is what you are differentiating with respect to and y is f(x)) y = a^x then dy/dx = lna.a^x Go ask a math teacher or look up exponential function differentiation on the internet for why.