Question

let ( f(x)=cx+ln(cos(x)) ). for what value of ( c ) is ( f(\frac{pi}{4}) = 5 )?

1. -/1 points

use logarithmic differentiation to find the derivative of the function.
( y = x^{7x} )

Explanation:

Step1: Differentiate $f(x)$

Using sum - rule and chain - rule, if $f(x)=cx+\ln(\cos(x))$, then $f'(x)=c+\frac{-\sin(x)}{\cos(x)}=c - \tan(x)$.

Step2: Substitute $x = \frac{\pi}{4}$ into $f'(x)$

Substitute $x=\frac{\pi}{4}$ into $f'(x)$. We know that $\tan(\frac{\pi}{4}) = 1$, so $f'(\frac{\pi}{4})=c - 1$.

Step3: Solve for $c$

Since $f'(\frac{\pi}{4}) = 5$, we set up the equation $c - 1=5$. Adding 1 to both sides gives $c=6$.

for second part:

Step1: Take the natural logarithm of both sides of $y = x^{7x}$

$\ln y=\ln(x^{7x})$. Using the power - rule of logarithms $\ln(a^{b})=b\ln(a)$, we get $\ln y = 7x\ln(x)$.

Step2: Differentiate both sides with respect to $x$

Using the product - rule $(uv)'=u'v + uv'$ where $u = 7x$ and $v=\ln(x)$. The derivative of $\ln y$ with respect to $x$ is $\frac{y'}{y}$, the derivative of $7x$ is 7, and the derivative of $\ln(x)$ is $\frac{1}{x}$. So $\frac{y'}{y}=7\ln(x)+7x\cdot\frac{1}{x}=7\ln(x)+7$.

Step3: Solve for $y'$

Multiply both sides by $y$. Since $y = x^{7x}$, then $y'=x^{7x}(7\ln(x)+7)$.

Answer:

$6$