How do you use the chain rule to differentiate #y=sec2x^4#?

Question

Charles Arocha · Accepted Answer

#dy/dx=8x^3sec(2x^4)tan(2x^4)# #"given "y=g(h(x))" then "# #dy/dx=g'(h(x))xxh'(x)# #y=sec(2x^4)# #rArrdy/dx=sec(2x^4)tan(2x^4)xxd/dx(2x^4)# #color(white)(rArrdy/dx)=8x^3sec(2x^4)tan(2x^4)#

Luke Alvarez · Answer

undefined To differentiate $ y = \sec^2(x^4) $ using the chain rule, first, recognize that $ \sec^2(u) $ is the derivative of $ \tan(u) $ with respect to $ u $. Then, differentiate the outer function with respect to the inner function (using the chain rule).

$ \frac{dy}{dx} = \frac{d}{dx}[\sec^2(x^4)] $

Let $ u = x^4 $.

$ \frac{dy}{du} = \frac{d}{du}[\sec^2(u)] $

Apply the chain rule:

$ \frac{dy}{du} = 2\sec(u)\tan(u) $

Now, differentiate $ u = x^4 $ with respect to $ x $:

$ \frac{du}{dx} = 4x^3 $

Finally, using the chain rule, multiply $ \frac{dy}{du} $ and $ \frac{du}{dx} $:

$ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} = 2\sec(x^4)\tan(x^4) \times 4x^3 $

$ \frac{dy}{dx} = 8x^3\sec(x^4)\tan(x^4) $

So, the derivative of $ y = \sec^2(x^4) $ is $ 8x^3\sec(x^4)\tan(x^4) $.