Assuming f is differentiable, find the derivatives of: 1°

Assuming $f$ is differentiable, find the derivatives of: 1° $y = f\bigl(x^{3}\bigr)$; 2° $y = f\bigl[f(x)\bigr]$; 3° $y = f\bigl(e^{x}\bigr)e^{f(x)}$; 4° $y = f(\ln x)$; 5° $y = \ln f(x)$.

Step 1: $y = f\bigl(x^{3}\bigr)$

Chain rule with inner function $u = x^{3}$, $u' = 3x^{2}$. Note that $f'$ is evaluated at $x^{3}$, not at $x$: $y' = f'\bigl(x^{3}\bigr)\cdot 3x^{2}$.

\[1^{\circ}\qquad y' = 3x^{2}f'\bigl(x^{3}\bigr)\]

Step 2: $y = f\bigl[f(x)\bigr]$

Chain rule with inner function $u = f(x)$, $u' = f'(x)$: $y' = f'\bigl[f(x)\bigr]\cdot f'(x)$.

The two factors are different: the first is $f'$ evaluated at $f(x)$, the second $f'$ evaluated at $x$.

\[2^{\circ}\qquad y' = f'\bigl[f(x)\bigr]\cdot f'(x)\]

Step 3: $y = f\bigl(e^{x}\bigr)\,e^{f(x)}$

A product, each factor itself a composite. With $u = f\bigl(e^{x}\bigr)$ and $v = e^{f(x)}$: $u' = f'\bigl(e^{x}\bigr)e^{x}, \qquad v' = e^{f(x)}f'(x)$.

The product rule then gives, after taking out the common factor $e^{f(x)}$: \[y' = f'\bigl(e^{x}\bigr)e^{x}e^{f(x)} + f\bigl(e^{x}\bigr)e^{f(x)}f'(x) = e^{f(x)}\Bigl[e^{x}f'\bigl(e^{x}\bigr) + f\bigl(e^{x}\bigr)f'(x)\Bigr].\]

\[3^{\circ}\qquad y' = e^{f(x)}\Bigl[e^{x}f'\bigl(e^{x}\bigr) + f\bigl(e^{x}\bigr)f'(x)\Bigr]\]

Step 4: $y = f(\ln x)$

Chain rule with $u = \ln x$, $u' = \dfrac1x$: $4^{\circ}\qquad y' = \dfrac{f'(\ln x)}{x}$.

Step 5: $y = \ln f(x)$

Chain rule with the logarithm outside, $u = f(x)$: $5^{\circ}\qquad y' = \dfrac{f'(x)}{f(x)}$.

Step 6: Verification

Taking $f(t) = t^{2}$, so $f'(t) = 2t$: \[1^{\circ}:\ y = x^{6},\ y' = 6x^{5}; \quad \text{formula: } 3x^{2}\cdot 2x^{3} = 6x^{5} \;\;\checkmark\]

\[2^{\circ}:\ y = x^{4},\ y' = 4x^{3}; \quad \text{formula: } 2x^{2}\cdot 2x = 4x^{3} \;\;\checkmark\]

\[5^{\circ}:\ y = \ln x^{2} = 2\ln|x|,\ y' = \tfrac{2}{x}; \quad \text{formula: } \tfrac{2x}{x^{2}} = \tfrac2x \;\;\checkmark\]