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Calculus 1 Finding the derivative
- Thread starter andy849
- Start date
Hello, andy849!
\(\displaystyle f'(x) \;=\;\dfrac{(x^2+1)^{\frac{1}{2}}\cdot 2\sec^2\!(x^3+2)\tan(x^3+2)3x^2 \;-\; \sec^2\!(x^3+2)\cdot\frac{1}{2}(x^2+1)^{-\frac{1}{2}}2x}{x^2+1}\)
\(\displaystyle f'(x) \;=\;\dfrac{6x^2(x^2+1)^{\frac{1}{2}}\sec^2\!(x^3+2)\tan(x^3+2) \;-\; x(x^2+1)^{-\frac{1}{2}}\sec^2\!(x^3+2)}{x^2+1}\)
Factor:
\(\displaystyle f'(x) \;=\;\dfrac{x(x^2+1)^{-\frac{1}{2}}\sec^2\!(x^3+2)\cdot\big[6x(x^2+1)\tan(x^3+2) \:-\: 1\big]}{x^2+1}\)
\(\displaystyle f'(x) \;=\;\dfrac{x\sec^2\!(x^3+2)\cdot \big[6x(x^2+1)\tan(x^3+2) - 1\big]}{(x^2+1)^{\frac{3}{2}}} \)
Quotient Rule:I'm getting stuck when I have to factor things out
. . \(\displaystyle f(x) \:=\:\dfrac{\sec^2\!(x^3+2)}{(x^2+1)^{\frac{1}{2}}}\)
\(\displaystyle f'(x) \;=\;\dfrac{(x^2+1)^{\frac{1}{2}}\cdot 2\sec^2\!(x^3+2)\tan(x^3+2)3x^2 \;-\; \sec^2\!(x^3+2)\cdot\frac{1}{2}(x^2+1)^{-\frac{1}{2}}2x}{x^2+1}\)
\(\displaystyle f'(x) \;=\;\dfrac{6x^2(x^2+1)^{\frac{1}{2}}\sec^2\!(x^3+2)\tan(x^3+2) \;-\; x(x^2+1)^{-\frac{1}{2}}\sec^2\!(x^3+2)}{x^2+1}\)
Factor:
\(\displaystyle f'(x) \;=\;\dfrac{x(x^2+1)^{-\frac{1}{2}}\sec^2\!(x^3+2)\cdot\big[6x(x^2+1)\tan(x^3+2) \:-\: 1\big]}{x^2+1}\)
\(\displaystyle f'(x) \;=\;\dfrac{x\sec^2\!(x^3+2)\cdot \big[6x(x^2+1)\tan(x^3+2) - 1\big]}{(x^2+1)^{\frac{3}{2}}} \)
Hello, andy849!
Quotient Rule:
\(\displaystyle f'(x) \;=\;\dfrac{(x^2+1)^{\frac{1}{2}}\cdot 2\sec^2\!(x^3+2)\tan(x^3+2)3x^2 \;-\; \sec^2\!(x^3+2)\cdot\frac{1}{2}(x^2+1)^{-\frac{1}{2}}2x}{x^2+1}\)
\(\displaystyle f'(x) \;=\;\dfrac{6x^2(x^2+1)^{\frac{1}{2}}\sec^2\!(x^3+2)\tan(x^3+2) \;-\; x(x^2+1)^{-\frac{1}{2}}\sec^2\!(x^3+2)}{x^2+1}\)
Factor:
\(\displaystyle f'(x) \;=\;\dfrac{x(x^2+1)^{-\frac{1}{2}}\sec^2\!(x^3+2)\cdot\big[6x(x^2+1)\tan(x^3+2) \:-\: 1\big]}{x^2+1}\)
\(\displaystyle f'(x) \;=\;\dfrac{x\sec^2\!(x^3+2)\cdot \big[6x(x^2+1)\tan(x^3+2) - 1\big]}{(x^2+1)^{\frac{3}{2}}} \)
Hi Soroban. There is a slight error in your derivative, In the first step, it should be 2sec(x^3+2) not 2sec²(x^3+2) and then that fix is needed throughout the rest of the steps.