What is the derivative of #x^sin(x)#?

Answer 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!

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#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]#.
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.
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

#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))#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 4

The derivative of ( x^{\sin(x)} ) with respect to ( x ) is given by ( x^{\sin(x)}(\ln(x)\cos(x)+\frac{\sin(x)}{x}) ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7