I have the following formula -
$$ f(x) = \left(0.1 e^{-1.5{x}^{0.2}} + 0.9 e^{-0.5{x}^{0.1}}\right)^{c}$$
where $\bf c$ is a constant value. How can I solve $f'(x)$ ?
According to the answer, I have got:
$$ f'(x) = c \left(0.1 e^{-1.5{x}^{0.2}} + 0.9 e^{-0.5{x}^{0.1}}\right) ^{c-1} (-0.03 e^{-1.5{x}^{0.2}} - 0.045 e^{-0.5{x}^{0.1}})$$
Is my answer right?
$[g(x)^c]' = c\cdot g(x)^{c-1}g'(x)$
I assume you can differentiate the term in the brackets (the $g(x)$).
As mentioned earlier, you have to use chain rule here. It is : $$\frac{d}{dx} \left(f(g(x)\right) = f'(g(x)) g'(x) -----\text{equation (1)} $$ The given expression $$ f(x) = \left( 0.1 e^{-1.5x^{0.2}} + 0.9 e^{-0.5 x^{0.1}} \right)^c $$ is quite similar to the equation (1) mentioned above.
In order to find $f'(x)$ , let us first make it a little bit simpler.
$$f(x) = \left(g(x)\right)^c \quad ; \text{where} \space g(x) = 0.1e^{-1.5x^{0.2}} + 0.9 e^{-0.5x^{0.1}} $$
Now, according to the chain rule : $$\frac{d}{dx} f(x) = c \times g'(x) \times g(x)^{c-1} $$
Now, you have to find $g'(x)$ as the other two terms will remain as it is. We don't need to touch them. To solve for $g'(x)$ , you will need to a little bit of work here. Remember that : $$\frac{d(a \times f(x) )}{dx} = a \frac{df(x)}{dx} $$ where a is a constant.
Good Luck with your problem. Let me know if you get any problems while working on it.
In order to differentiate this formula, you need to be familiar with the chain rule. It says that: $$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$$
Hence, the derivative of your formula becomes:
$$c\cdot (0.1e^{-1.5x^{0.2}}+0.5e^{-0.5x^{0.1}})^{c-1}\cdot \frac{d}{dx}(0.1e^{-1.5x^{0.2}}+0.5e^{-0.5x^{0.1}})$$
In the second part of the derivative, you need to apply the chain rule again. The solution becomes:
$c\cdot (0.1e^{-1.5x^{0.2}}+0.5e^{-0.5x^{0.1}})^{c-1}\cdot(-\frac{0.045e^{-0.5x^{0.1}}}{x^{0.9}}-\frac{0.03e^{-1.5x^{0.2}}}{x^{0.8}})$.
