Home > Calculus > Implicit Differentiation

How do you differentiate #e^y = (tanx)^sinx#?

Answer 1

Isabella Ackerman

#y' = cosxln(tanx) + secx#

Take the natural logarithm of both sides.

#ln(e^y) = ln(tanx)^sinx#

#ylne = sinxln(tanx)#

#y = sinxln(tanx)#

This is now a normal function explicitly defined by #y#. The derivative can be found by using the product and chain rules.

Let #y = g(x)h(x)#, with #g(x) = sinx# and #h(x) = ln(tanx)#. The derivative of #g(x)#, by first principles, in #g'(x) = cosx#.

However, we must use the chain rule to find #h'(x)#. We let #y = lnu# and #u = tanx#. Then #dy/(du) = 1/u# and #(du)/dx = sec^2x#. This means that #h'(x) = 1/u * sec^2x = sec^2x/tanx = (1/cos^2x)/(sinx/cosx) = secxcscx#

We now apply the product rule, which states that #y' = g'(x)h(x) + h'(x)g(x)#.

#y' = cosx(ln(tanx)) + sinx(secxcscx)#

#y' = cosxln(tanx) + secx#

Hopefully this helps!

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

Answer 2

Emily Ammerman

To differentiate ( e^y = (\tan x)^{\sin x} ):

Take the natural logarithm of both sides to eliminate the exponential function.
Apply implicit differentiation.
Solve for ( \frac{dy}{dx} ).

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

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.