Resourcesdown chevron
Testimonials
Plans
Sign in
Sign up
Resourcesright arrow
Testimonials
Plans
Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Let h(x)=ln(x)cos(x). h^(')(x)=

Let h(x)=ln⁡(x)cos⁡(x) h(x)=\ln (x) \cos (x) .\newlineh′(x)= h^{\prime}(x)=

View step-by-step helpView step-by-step help
Homeright chevron icon
Math Problemsright chevron icon
Algebra 2right chevron icon
Evaluate functionsright chevron icon

Full solution

Q. Let h(x)=ln⁡(x)cos⁡(x) h(x)=\ln (x) \cos (x) .\newlineh′(x)= h^{\prime}(x)=
  1. Use Product Rule: We need to use the product rule to differentiate h(x)=ln⁡(x)cos⁡(x)h(x) = \ln(x)\cos(x), which states that (fg)′=f′g+fg′(fg)' = f'g + fg'. Let's differentiate ln⁡(x)\ln(x) and cos⁡(x)\cos(x) separately.
  2. Differentiate ln⁡(x)\ln(x): Differentiate ln⁡(x)\ln(x) to get 1x\frac{1}{x}.
  3. Differentiate cos⁡(x)\cos(x): Differentiate cos⁡(x)\cos(x) to get −sin⁡(x)-\sin(x).
  4. Apply Product Rule: Now apply the product rule: h′(x)=(ln⁡(x))′(cos⁡(x))+(ln⁡(x))(cos⁡(x))′h'(x) = (\ln(x))'(\cos(x)) + (\ln(x))(\cos(x))'.
  5. Substitute Derivatives: Substitute the derivatives into the product rule: h′(x)=(1x)(cos⁡(x))+ln⁡(x)(−sin⁡(x))h'(x) = (\frac{1}{x})(\cos(x)) + \ln(x)(-\sin(x)).
  6. Simplify Expression: Simplify the expression: h′(x)=cos⁡(x)x−ln⁡(x)sin⁡(x)h'(x) = \frac{\cos(x)}{x} - \ln(x)\sin(x).

More problems from Evaluate functions

Question
We want to factor the following expression:\newlinex4+9 x^{4}+9 \newlineWhich pattern can we use to factor the expression?\newlineU U and V V are either constant integers or single-variable expressions.\newlineChoose 11 answer:\newline(A) (U+V)2 (U+V)^{2} or (U−V)2 (U-V)^{2} \newline(B) (U+V)(U−V) (U+V)(U-V) \newline(C) We can't use any of the patterns.
Get tutor helpright-arrow

Posted 3 months ago

Question
z=−19+3.14i z=-19+3.14 i \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=−19 and Im⁡(z)=3.14 \begin{array}{l} \operatorname{Re}(z)=-19 \text { and } \\ \operatorname{Im}(z)=3.14 \end{array} \newline(B)\newlineRe⁡(z)=3.14i and Im⁡(z)=−19 \begin{array}{l} \operatorname{Re}(z)=3.14 i \text { and } \\ \operatorname{Im}(z)=-19 \end{array} \newline(c)\newlineRe⁡(z)=−19 and Im⁡(z)=3.14i \begin{array}{l} \operatorname{Re}(z)=-19 \text { and } \\ \operatorname{Im}(z)=3.14 i \end{array} \newline(D)\newlineRe⁡(z)=3.14 and Im⁡(z)=−19 \begin{array}{l} \operatorname{Re}(z)=3.14 \text { and } \\ \operatorname{Im}(z)=-19 \end{array}
Get tutor helpright-arrow

Posted 3 months ago

Question
z=−60−12iRe⁡(z)=Im⁡(z)= \begin{array}{l}z=-60-12 i \\ \operatorname{Re}(z)= \\ \operatorname{Im}(z)=\end{array}
Get tutor helpright-arrow

Posted 3 months ago

Question
z=−45i−15.5 z=-45 i-15.5 \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=−45 and Im⁡(z)=−15.5 \begin{array}{l} \operatorname{Re}(z)=-45 \text { and } \\ \operatorname{Im}(z)=-15.5 \end{array} \newline(B)\newlineRe⁡(z)=−15.5 and Im⁡(z)=−45 \begin{array}{l} \operatorname{Re}(z)=-15.5 \text { and } \\ \operatorname{Im}(z)=-45 \end{array} \newline(C)\newlineRe⁡(z)=−45i and Im⁡(z)=−15.5 \begin{array}{l} \operatorname{Re}(z)=-45 i \text { and } \\ \operatorname{Im}(z)=-15.5 \end{array} \newline(D)\newlineRe⁡(z)=−15.5 and Im⁡(z)=−45i \begin{array}{l} \operatorname{Re}(z)=-15.5 \text { and } \\ \operatorname{Im}(z)=-45 i \end{array}
Get tutor helpright-arrow

Posted 3 months ago

Question
z=−12i+11 z=-12 i+11 \newlineWhat are the real and imaginary parts of z z ?\newlineChoose 11 answer:\newline(A)\newlineRe⁡(z)=11 and Im⁡(z)=−12 \begin{array}{l} \operatorname{Re}(z)=11 \text { and } \\ \operatorname{Im}(z)=-12 \end{array} \newline(B)\newlineRe⁡(z)=11 and Im⁡(z)=−12i \begin{array}{l} \operatorname{Re}(z)=11 \text { and } \\ \operatorname{Im}(z)=-12 i \end{array} \newline(C)\newlineRe⁡(z)=−12i and Im⁡(z)=11 \begin{array}{l} \operatorname{Re}(z)=-12 i \text { and } \\ \operatorname{Im}(z)=11 \end{array} \newline(D)\newlineRe⁡(z)=−12 and Im⁡(z)=11 \begin{array}{l} \operatorname{Re}(z)=-12 \text { and } \\ \operatorname{Im}(z)=11 \end{array}
Get tutor helpright-arrow

Posted 3 months ago

Question
(18+4i)+(−11+23i)= (18+4 i)+(-11+23 i)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 3 months ago

Question
(14+60i)+(−30+2i)= (14+60 i)+(-30+2 i)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 3 months ago

Question
(3−91i)+(67)= (3-91 i)+(67)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 3 months ago

Question
(−71+2i)+(88−12i)= (-71+2 i)+(88-12 i)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 3 months ago

Question
(−33−2i)−(50+9i)= (-33-2 i)-(50+9 i)= \newlineExpress your answer in the form (a+bi) (a+b i) .
Get tutor helpright-arrow

Posted 3 months ago

View all (15)