# find y' for y=e^(sin(42x)) use the 3 fold chain rule if you can

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### 2 Answers

You should remember that you need to use the chain rule when you should evaluate the derivative of composition of functions such that:

`(f(g(x)))' = f'(g(x))*g'(x)*x'`

Reasoning by analogy yields:

`(dy)/(dx) = (d(e^(sin(42x))))/(dx)`

`(dy)/(dx) = (d(e^(sin(42x))))/(dx)*(d(sin (42 x)))/(dx)*(d(42x))/(dx)`

`(dy)/(dx) = e^(sin(42x))*cos(42x)*42`

**Hence, evaluating the derivative of the given composition of functions, using the chain rule, yields `(dy)/(dx) = 42e^(sin(42x))*cos(42x).` **

Three fold chain rule can be applied here if we think

y = f(g(h(x))) = e^(Sin(42x))

where, "f" is exponential, "g" is Sine, and "h" is 42*x,

so

y' = dy/dx = (df/dg)(dg/dh)(dh/dx) ----------------(1)

now, f(g) = e^g

so df/dg = e^g

then, g(h) = Sin(h)

so dg/dh = Cos(h)

finally, h(x) = 42*x

so dh/dx = 42

Hence,

**using equation no. (1) we get,**

**y' = e^g * Cos(h) * 42**

** = e^(Sin(42x)) * Cos(42x) * 42, is the answer.**