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- Find the derivative using the definition
- Integrate by partial fractions
- Product of Binomials with Common Term
- FOIL Method
- Integrate by substitution
- Integrate by parts
- Integrate using tabular integration
- Integrate by trigonometric substitution
- Weierstrass Substitution
- Prove from LHS (left-hand side)
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Find the derivative of $\arctan\left(\frac{\sin\left(\frac{3}{25}\right)+\frac{1}{5}\sin\left(\frac{16}{25}\right)}{\cos\left(\frac{3}{25}\right)-\frac{1}{5}\cos\left(\frac{16}{25}\right)}\right)$ using the definition. Apply the definition of the derivative: $\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. The function $f(x)$ is the function we want to differentiate, which is $\arctan\left(\frac{\sin\left(\frac{3}{25}\right)+\frac{1}{5}\sin\left(\frac{16}{25}\right)}{\cos\left(\frac{3}{25}\right)-\frac{1}{5}\cos\left(\frac{16}{25}\right)}\right)$. Substituting $f(x+h)$ and $f(x)$ on the limit, we get
Learn how to solve definition of derivative problems step by step online.
$\lim_{h\to0}\left(\frac{\arctan\left(\frac{\sin\left(\frac{3}{25}\right)+\frac{1}{5}\sin\left(\frac{16}{25}\right)}{\cos\left(\frac{3}{25}\right)-\frac{1}{5}\cos\left(\frac{16}{25}\right)}\right)-\arctan\left(\frac{\sin\left(\frac{3}{25}\right)+\frac{1}{5}\sin\left(\frac{16}{25}\right)}{\cos\left(\frac{3}{25}\right)-\frac{1}{5}\cos\left(\frac{16}{25}\right)}\right)}{h}\right)$
Learn how to solve definition of derivative problems step by step online. Find the derivative of arctan((sin(3/25)+1/5sin(16/25))/(cos(3/25)-1/5cos(16/25))) using the definition. Find the derivative of \arctan\left(\frac{\sin\left(\frac{3}{25}\right)+\frac{1}{5}\sin\left(\frac{16}{25}\right)}{\cos\left(\frac{3}{25}\right)-\frac{1}{5}\cos\left(\frac{16}{25}\right)}\right) using the definition. Apply the definition of the derivative: \displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}. The function f(x) is the function we want to differentiate, which is \arctan\left(\frac{\sin\left(\frac{3}{25}\right)+\frac{1}{5}\sin\left(\frac{16}{25}\right)}{\cos\left(\frac{3}{25}\right)-\frac{1}{5}\cos\left(\frac{16}{25}\right)}\right). Substituting f(x+h) and f(x) on the limit, we get. Cancel like terms \arctan\left(\frac{\sin\left(\frac{3}{25}\right)+\frac{1}{5}\sin\left(\frac{16}{25}\right)}{\cos\left(\frac{3}{25}\right)-\frac{1}{5}\cos\left(\frac{16}{25}\right)}\right) and -\arctan\left(\frac{\sin\left(\frac{3}{25}\right)+\frac{1}{5}\sin\left(\frac{16}{25}\right)}{\cos\left(\frac{3}{25}\right)-\frac{1}{5}\cos\left(\frac{16}{25}\right)}\right). Zero divided by anything is equal to zero. The limit of a constant is just the constant.