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Question

Question: What is the derivative of \(\tan xy?\)...

What is the derivative of tanxy?\tan xy?

Explanation

Solution

We will use the usual rule of differentiation of functions. If we need to differentiate a function y=f(u)y=f\left( u \right) with respect to x,x, then we will first differentiate the whole of the function with respect to uu and then we will differentiate uu with respect to x.x. That can be mathematically expressed as df(u)dx=dfdududx.\dfrac{df\left( u \right)}{dx}=\dfrac{df}{du}\dfrac{du}{dx}.

Complete step-by-step solution:
Let us consider the given trigonometric function tanxy.\tan xy.
We are asked to find the derivative of the given trigonometric function.
We know that if uu is a function of xx and ff is a function of uu and if we are asked to find the derivative of the function ff with respect to x,x, then we will have to differentiate ff with respect to uu and then multiply the derivative with the derivative of uu with respect to x.x.
And we can express it mathematically as df(u)dx=dfdududx.\dfrac{df\left( u \right)}{dx}=\dfrac{df}{du}\dfrac{du}{dx}.
Now, let us compare the given function with the above identity.
Then, we will get f(u)=tanxyf\left( u \right)=\tan xy and u=xy.u=xy.
When we differentiate the given function with the help of the above identity, we will get the left-hand side of the equation as df(u)dx=dtanxydx.\dfrac{df\left( u \right)}{dx}=\dfrac{d\tan xy}{dx}.
Now let us take u=xyu=xy and we will get dfdx=ddxtanu\dfrac{df}{dx}=\dfrac{d}{dx}\tan u and dudx=dxydx.\dfrac{du}{dx}=\dfrac{dxy}{dx}.
We know that the derivative of tanu\tan u with respect to uu is sec2u{{\sec }^{2}}u and the derivative of xyxy with respect to xx is y.y.
That is, we will get ddxtanu=sec2u\dfrac{d}{dx}\tan u={{\sec }^{2}}u and dxydx=y.\dfrac{dxy}{dx}=y.
Therefore, we will get the right-hand side of the identity as dfdududx=ddutanududx=ddutanudxydx.\dfrac{df}{du}\dfrac{du}{dx}=\dfrac{d}{du}\tan u\dfrac{du}{dx}=\dfrac{d}{du}\tan u\dfrac{dxy}{dx}.
Now, we will get dfdududx=ddutanududx=sec2uy.\dfrac{df}{du}\dfrac{du}{dx}=\dfrac{d}{du}\tan u\dfrac{du}{dx}={{\sec }^{2}}u\cdot y.
And that is. dfdududx=ysec2xy.\dfrac{df}{du}\dfrac{du}{dx}=y{{\sec }^{2}}xy.
Hence the derivative of the given trigonometric function is ysec2xy.y{{\sec }^{2}}xy.

Note: When we differentiate a function, we should always remember the identity df(u)dx=dfdududx.\dfrac{df\left( u \right)}{dx}=\dfrac{df}{du}\dfrac{du}{dx}. We sometimes make a mistake by forgetting the part dudx\dfrac{du}{dx} in the identity and it leads us to the wrong answer.