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h(x)=(arctan(x^(5)+2x))^(7)

h(x)=(arctan⁡(x5+2x))7 h(x)=\left(\arctan \left(x^{5}+2 x\right)\right)^{7}

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Full solution

Q. h(x)=(arctan⁡(x5+2x))7 h(x)=\left(\arctan \left(x^{5}+2 x\right)\right)^{7}
  1. Use Chain Rule: step_1: Use the chain rule to differentiate h(x)=(arctan⁡(x5+2x))7h(x) = (\arctan(x^{5}+2x))^{7}. Let u=arctan⁡(x5+2x)u = \arctan(x^{5}+2x), then h(x)=u7h(x) = u^{7}. dhdu=7u6\frac{dh}{du} = 7u^{6}.
  2. Find dudx\frac{du}{dx}: step_2: Now find dudx\frac{du}{dx}, which is the derivative of u=arctan⁡(x5+2x)u = \arctan(x^{5}+2x). dudx=11+(x5+2x)2∗ddx(x5+2x)\frac{du}{dx} = \frac{1}{1+(x^{5}+2x)^2} * \frac{d}{dx}(x^{5}+2x).
  3. Differentiate Inside Function: step_3: Differentiate the inside function x5+2xx^{5}+2x.ddx(x5+2x)=5x4+2\frac{d}{dx}(x^{5}+2x) = 5x^{4}+2.
  4. Combine Results: step_44: Combine the results from step_22 and step_33. dudx=5x4+21+(x5+2x)2\frac{du}{dx} = \frac{5x^{4}+2}{1+(x^{5}+2x)^{2}}.
  5. Apply Chain Rule: step_5: Apply the chain rule by multiplying dhdu\frac{dh}{du} from step_1 with dudx\frac{du}{dx} from step_4.\newlinef′(x)=dhdx=dhdu×dudx=7u6×(5x4+2)(1+(x5+2x)2)f'(x) = \frac{dh}{dx} = \frac{dh}{du} \times \frac{du}{dx} = 7u^6 \times \frac{(5x^{4}+2)}{(1+(x^{5}+2x)^2)}.\newlineBut we made a mistake, it should be 7u6×dudx7u^6 \times \frac{du}{dx} without the (5x4+2)(5x^{4}+2).

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