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Solving: $\frac{d}{dx}\left(y=\left(2x+1\right)^{\sin\left(x\right)}\right)$
Solve instead: $\frac{dy}{dx}\left(y=\left(2x+1\right)^{\sin\left(x\right)}\right)$

Exercise

$\frac{dy}{dx}\left(y=\left(2x+1\right)^{\sin\left(x\right)}\right)$

Step-by-step Solution

1

Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dx}\left(y\right)=\frac{d}{dx}\left(\left(2x+1\right)^{\sin\left(x\right)}\right)$
2

The derivative of the linear function is equal to $1$

$y^{\prime}=\frac{d}{dx}\left(\left(2x+1\right)^{\sin\left(x\right)}\right)$
3

The derivative $\frac{d}{dx}\left(\left(2x+1\right)^{\sin\left(x\right)}\right)$ results in $\left(\cos\left(x\right)\ln\left(2x+1\right)+\frac{2\sin\left(x\right)}{2x+1}\right)\left(2x+1\right)^{\sin\left(x\right)}$

$y^{\prime}=\left(\cos\left(x\right)\ln\left(2x+1\right)+\frac{2\sin\left(x\right)}{2x+1}\right)\left(2x+1\right)^{\sin\left(x\right)}$

Final answer to the exercise

$y^{\prime}=\left(\cos\left(x\right)\ln\left(2x+1\right)+\frac{2\sin\left(x\right)}{2x+1}\right)\left(2x+1\right)^{\sin\left(x\right)}$

Try other ways to solve this exercise

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  • Find the derivative using the definition
  • Find the derivative using the product rule
  • Find the derivative using the quotient rule
  • Find the derivative using logarithmic differentiation
  • Find the derivative
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Integrate by substitution
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