Question 9 Let f(x)=tanh(x2+1)f(x) = \tanh(x^2 + 1). (a) Find the derivative f′(x)f'(x) using the chain rule and hyperbolic identities. (b) Determine the value of the derivative at x=0x = 0. Show solutionHide solution+Question 9 - Solution We are given: f(x)=tanh(x2+1)f(x) = \tanh(x^2 + 1) (a) Find f′(x)f'(x): Use the chain rule: f′(x)=ddx[tanh(u)]⋅ddx(x2+1)f'(x) = \frac{d}{dx} \left[ \tanh(u) \right] \cdot \frac{d}{dx} (x^2 + 1) Recall: ddx[tanh(u)]=sech2(u)\frac{d}{dx} [\tanh(u)] = \text{sech}^2(u) So let u=x2+1u = x^2 + 1, then: f′(x)=sech2(x2+1)⋅2xf'(x) = \text{sech}^2(x^2 + 1) \cdot 2x f′(x)=2x⋅sech2(x2+1)\boxed{f'(x) = 2x \cdot \text{sech}^2(x^2 + 1)} (b) Evaluate at x=0x = 0: f′(0)=2⋅0⋅sech2(02+1)=0f'(0) = 2 \cdot 0 \cdot \text{sech}^2(0^2 + 1) = 0 f′(0)=0\boxed{f'(0) = 0}
