Find d by d𝑥 of inverse sin of 𝑥 over 𝑎, where 𝑎 is not equal to zero.
In this question, we need to differentiate the inverse or arcsin of 𝑥 over 𝑎 with respect to 𝑥. We will begin this question by letting 𝑦 equal the inverse sin of 𝑥 over 𝑎. If we take the sin of both sides of this equation, we get sin 𝑦 is equal to 𝑥 over 𝑎. Our next step is to differentiate both sides of this equation with respect to 𝑥. We know that differentiating sin 𝑥 gives us cos 𝑥. This means that differentiating sin 𝑦 with respect to 𝑥 using our knowledge of implicit differentiation gives us cos 𝑦 multiplied by d𝑦 by d𝑥. Differentiating the right-hand side gives us one over the constant 𝑎. We can then divide both sides of this equation by cos 𝑦 such that d𝑦 by d𝑥 is equal to one over 𝑎 multiplied by cos 𝑦.
We now have an expression for d𝑦 by d𝑥; however, this is not given in terms of 𝑥. If we go back to the equation sin 𝑦 is equal to 𝑥 over 𝑎, squaring both sides of this equation gives us sin squared 𝑦 is equal to 𝑥 squared over 𝑎 squared. One of our trigonometric identities states that sin squared 𝜃 plus cos squared 𝜃 is equal to one. Rearranging this, we see that sin squared 𝜃 is equal to one minus cos squared 𝜃. This means that sin squared 𝑦 is equal to one minus cos squared 𝑦. We now have one minus cos squared 𝑦 is equal to 𝑥 squared divided by 𝑎 squared. Multiplying both sides by 𝑎 squared gives us 𝑎 squared minus 𝑎 squared cos squared 𝑦 is equal to 𝑥 squared.
We can then rearrange this equation so that 𝑎 squared minus 𝑥 squared is equal to 𝑎 squared multiplied by cos squared 𝑦. Finally, we square root both sides of this equation. The right-hand side becomes 𝑎 cos 𝑦. And the left-hand side is the square root of 𝑎 squared minus 𝑥 squared. d𝑦 by d𝑥 is therefore equal to one over the square root of 𝑎 squared minus 𝑥 squared. This is the derivative of inverse sin of 𝑥 over 𝑎.