Find the derivative of the following function

Question

Given function:

$$f(x) = e^{\sin(x^2)}$$

Find the derivative of the above function where '$e$' stands for some constant.

I assume the correct answer is $e^{\sin(x^2)}\cdot \cos(x^2)\cdot 2x$, which method is best to solve like a question this?? Please help out to resolve this problem

Answer 1

Use the chain rule, $$(f(g(x))'=f'(g(x))g'(x)$$

For three functions, $$(f(g(h(x))))'=f'(g(h(x)))g'(h(x))h'(x)$$

Set $f(x) = e^x, g(x) = \sin{x}, h(x)=x^2$ to arrive at the desired answer.

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Calculations: $f'(x)=e^x, g'(x)=\cos{x}, h'(x)=2x$, thus, the result is $$e^{\sin{x^2}}\cdot \cos{x^2}\cdot 2x$$

Answer 2

Use the following rule

$$f(x)=e^{h(x)} \implies f'(x)=\frac {df}{dx}=\frac {de^{h(x)}}{dh(x)}\frac {dh(x)}{dx}=e^{h(x)}\frac {dh(x)}{dx}=e^{h(x)}\times h'(x)$$

Answer 3

$$y=a^u \implies y' = u' a^u \ln a$$

In your case $$u= \sin (x^2)$$ so $$u'=2x \cos(x^2)$$

$$ y'= 2x \cos(x^2)e^{\sin (x^2)} \ln e $$

If this $e$ is the well-known Euler' constant then $\ln e=1$ and you get

$$ y'= 2x \cos(x^2)e^{\sin (x^2)} $$

Answer 4

Besides getting the solution to your problem, take note that the general rule for $e^x$ is the same as $2^x$ except the outcome is different.

$f(x) = e^x, f'(x) = ln(e)(e^x)(1) = e^x$

Notice the $ln(e) = 1$ term disappears, as does the other $1$.

$f(x) = 2^x, f'(x) = ln(2)(2^x)(1) = ln(2)2^x$

Everyone tends to short-cut the general rule when dealing with derivatives of some form of $e^x$, but it is useful to remember it for dealing with other base numbers.