Question Video: Differentiating a Combination of Trigonometric Functions | Nagwa Question Video: Differentiating a Combination of Trigonometric Functions | Nagwa

# Question Video: Differentiating a Combination of Trigonometric Functions Mathematics • Third Year of Secondary School

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Find dπ¦/dπ₯, given that π¦ = β9 tan 6π₯ β csc 7π₯.

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### Video Transcript

Find dπ¦ by dπ₯, given that π¦ equals negative nine tan six π₯ minus csc seven π₯.

Looking at the function weβve been asked to differentiate, we can see that it consists of a trigonometric function, tan of six π₯, and also a reciprocal trigonometric function, csc of seven π₯, which we recall is equal to one over sin of seven π₯. In order to find each of these derivatives then, weβre going to need to recall the standard results for differentiating each of these functions. Firstly, the derivative with respect to π₯ of tan of ππ₯ for some constant π is equal to π multiplied by sec squared of ππ₯. We can see this if we recall that tan of ππ₯ is equal to sin of ππ₯ over cos ππ₯, and then we could apply the quotient rule to find this derivative, applying the standard results for differentiating sine and cosine functions.

We also recall that the derivative with respect to π₯ of csc of ππ₯ is equal to negative π multiplied by csc ππ₯ multiplied by cot ππ₯. And again, we can see this if we think of csc ππ₯ as one over sin ππ₯, and then we apply the quotient rule using our function π’ in the numerator as one and our function π£ in the denominator as sin ππ₯. Whilst itβs important to know how to derive each of these results, itβs also really helpful to commit them to memory because we can just quote them and then apply them in situations such as these.

Letβs find now an expression for dπ¦ by dπ₯ then. Applying the first rule, we have that the derivative of negative nine tan six π₯ is equal to negative nine multiplied by six sec squared six π₯. And applying the second rule, we have that the derivative of csc seven π₯ will be negative seven csc seven π₯ cot seven π₯. So, our entire expression for dπ¦ by dπ₯ is negative nine multiplied by six sec squared six π₯ minus negative seven csc seven π₯ cot seven π₯. We can simplify the coefficients to negative 54 and positive seven and then reorder the term so that the positive term comes first, if we wish. We find then that dπ¦ by dπ₯ is equal to seven csc seven π₯ cot seven π₯ minus 54 sec squared six π₯.

Remember, we need to know how to derive each of these results using the quotient rule, but itβs also helpful to commit them to memory.

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