Find the roots of $\ln\left(x\right)\left(\sec\left(x^2+x^3\right)+\tan\left(x\right)\right)$

Step-by-step Solution

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ln
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acsch
Solving: $\ln\left(x\right)\left(\sec\left(x^2+x^3\right)+\tan\left(x\right)\right)$

Final answer to the problem

$x=1$
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Step-by-step Solution

How should I solve this problem?

  • Find the roots
  • Solve for x
  • Find the derivative using the definition
  • Solve by quadratic formula (general formula)
  • Simplify
  • Find the integral
  • Find the derivative
  • Factor
  • Factor by completing the square
  • Find break even points
  • Load more...
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1

Find the roots of the polynomial $\ln\left(x\right)\left(\sec\left(x^2+x^3\right)+\tan\left(x\right)\right)$ by putting it in the form of an equation and then set it equal to zero

$\ln\left(x\right)\left(\sec\left(x^2+x^3\right)+\tan\left(x\right)\right)=0$

Learn how to solve implicit differentiation problems step by step online.

$\ln\left(x\right)\left(\sec\left(x^2+x^3\right)+\tan\left(x\right)\right)=0$

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Learn how to solve implicit differentiation problems step by step online. Find the roots of ln(x)(sec(x^2+x^3)+tan(x)). Find the roots of the polynomial \ln\left(x\right)\left(\sec\left(x^2+x^3\right)+\tan\left(x\right)\right) by putting it in the form of an equation and then set it equal to zero. Break the equation in 2 factors and set each factor equal to zero, to obtain simpler equations. Solve the equation (1). Take the variable outside of the logarithm.

Final answer to the problem

$x=1$

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Function Plot

Plotting: $\ln\left(x\right)\left(\sec\left(x^2+x^3\right)+\tan\left(x\right)\right)$

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1
2
3
4
5
6
7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Implicit Differentiation

Implicit differentiation makes use of the chain rule to differentiate implicitly defined functions. For differentiating an implicit function y(x), defined by an equation R(x, y) = 0, it is not generally possible to solve it explicitly for y(x) and then differentiate. Instead, one can differentiate R(x, y) with respect to x and y and then solve a linear equation in dy/dx for getting explicitly the derivative in terms of x and y.

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