What is the derivative of #x^sin(x)#?
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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
#dy/dx = x^sinx(cosxlnx+sinx/x)#
#y = x^sinx#
Take the natural logarithm of both sides.
#lny = ln(x^sinx)#
Use laws of logarithms to simplify.
#lny = sinxlnx#
Use the product rule and implicit differentiation to differentiate.
#1/y(dy/dx) = cosx(lnx) + 1/x(sinx)#
#1/y(dy/dx) = cosxlnx + sinx/x#
#dy/dx = (cosxlnx + sinx/x)/(1/y)#
#dy/dx = x^sinx(cosxlnx+sinx/x)#
Hopefully this helps!
#d/dx x^(sin x)=x^(sin x)[cos x * ln x + (sin x)/x]#.
When we have a function of #x# like #y=x^sin x#, where a single term contains #x# in both its base and its power, perhaps the easiest way to find the function's derivative is to first take the (natural) logarithm of both sides:
#ln y = ln (x^(sin x))#
#color(white)(ln y)=sin x * ln x#
This places all the #x#'s on the same "level". Then, take the derivative of both sides with respect to #x#:
#=>d/dx (ln y)=d/dx (sin x * ln x)#
Remembering that #y# is a function of #x#, we get
#=> 1/y*dy/dx=cos x * ln x + sin x (1/x)#
#=> color(white)"XXi"dy/dx=y[cos x * ln x + (sin x) /x]#
Since we began with #y=x^(sin x)#, we substitute this back in for #y# to get
#=> color(white)"XXi"dy/dx=x^(sin x)[cos x * ln x + (sin x)/x]#.
Note:
When #f(x)=g(x)^(h(x))#, you'll almost always see #g(x)^(h(x))# appear in the derivative of #f(x)#. If you don't, go back and double check your work to make sure things were done right.
#dy/dx = ((cos(x))ln(x) + (sin(x))/x)x^(sin(x))#
Given: #y = x^(sin(x))#
Use logarithmic differentiation.
#ln(y) = ln(x^(sin(x)))#
On the right side, use a property of logarithms, #ln(a^b) = (b)ln(a)#:
#ln(y) = (sin(x))ln(x)#
Use implicit differentiation on the left side:
#(dln(y))/dx = 1/ydy/dx#
Use the product rule on the right sides:
#(d(uv))/dx = (u')(v) + (u)(v')#
let #u = sin(x)#, then #u' = cos(x), v =ln(x), and v' = 1/x#
Substituting into the product rule:
#(d((sin(x))ln(x)))/dx = (cos(x))ln(x) + (sin(x))/x#
Put the equation back together:
#1/ydy/dx = (cos(x))ln(x) + (sin(x))/x#
Multiply both sides by y:
#dy/dx = ((cos(x))ln(x) + (sin(x))/x)y#
Substitute #x^(sin(x))# for y:
#dy/dx = ((cos(x))ln(x) + (sin(x))/x)x^(sin(x))#
