How do you find the first derivative of #y=(lnx)^tanx#?

Answer 1

Take the natural logarithm of both sides.

#ln[y] = ln[(lnx)^(tanx)]#
Simplify the right-hand side using the rule #ln[a^n} = nlna#:
#ln[y] = tanx[ln(lnx)]#

Differentiate both sides.

#1/y(dy/dx) = square#
Inset: #square#
The derivative of the right hand side is fairly complex. We have to first find the derivative of #ln(lnx)# using the chain rule, followed by the derivative of #tanx xx ln(lnx)#, using the product rule.
Let #y = ln(u)# and #u = lnx#. Then #y' = 1/u# and #u' = 1/x#.
#u' = 1/u xx 1/x = 1/(lnx) xx 1/x = 1/(xlnx)#
Now we have to find the derivative for #tanx# so that we have enough information to apply the product rule.
#tanx = sinx/cosx#

By the quotient rule:

#(tanx)' = (cosx xx cosx - (-sinx xx sinx))/(cosx)^2#
#(tanx)' = (cos^2x + sin^2x)/(cos^2x)#
Applying the pythagorean identity #cos^2x + sin^2x = 1#:
#(tanx)' = 1/(cos^2x) = sec^2x#

Next, apply the product rule:

Let #y = g(x) xx h(x)#, where #g(x) = ln(lnx) and h(x) = tanx#:
#y' = (1/(xlnx) xx tanx + ln(lnx) xx sec^2x)#
#y' = tanx/(xlnx) + sec^2xln(lnx)#
Putting this into the place of the #square#:
#1/y(dy/dx) = tanx/(xlnx) + sec^2xln(lnx)#
#dy/dx = ( tanx/(xlnx) + sec^2xln(lnx))/(1/y)#
#dy/dx = y( tanx/(xlnx) + sec^2xln(lnx))#
#dy/dx= (lnx)^(tanx)( tanx/(xlnx) + sec^2xln(lnx))#

I know it's quite messy, but it worked!!

Practice exercises:

a) #f(x) = log_2(sin(3x + 4))#
b) #g(x) = (2lnx)^(sec(3x + 5))#

Hopefully this helps, and good luck!

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

To find the first derivative of ( y = (\ln x)^{\tan x} ), we use the chain rule. The chain rule states that if we have a composite function ( f(g(x)) ), then the derivative is ( f'(g(x)) \cdot g'(x) ).

  1. Let ( u = \ln x ) and ( v = \tan x ).
  2. Rewrite ( y ) as ( y = u^v ).
  3. Find the derivatives ( \frac{du}{dx} ) and ( \frac{dv}{dx} ).
  4. Apply the chain rule: ( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} + \frac{dy}{dv} \cdot \frac{dv}{dx} ).
  5. Substitute ( \frac{dy}{du} ) and ( \frac{dy}{dv} ).
  6. Simplify the expression to find ( \frac{dy}{dx} ).
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